Heat Class 10th Solutions | Heat SSC Class 10 Questions And Answers
Fill in the blank and rewrite the sentence.
The amount of water vapour in air is determined in terms of its humidity.
Fill in the blank and rewrite the sentence.
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacity.
Fill in the blank and rewrite the sentence.
During transformation of liquid phase to solid phase, the latent heat is released.
Observe the following graph. Considering the change in volume of water as its temperature is raised from 0°C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?
Scientific and Written Exam Answer:
Water exhibits a unique behaviour known as anomalous expansion. Unlike most substances, which expand when heated and contract when cooled, water behaves differently between 0°C and 4°C.
Explanation of the graph:
- At 0°C: Water is in the form of ice, and its volume is relatively high due to the open structure of ice crystals.
- From 0°C to 4°C: Instead of expanding, water contracts, meaning its volume decreases as temperature increases.
- At 4°C: Water reaches its minimum volume and maximum density.
- Beyond 4°C: Water starts expanding like other liquids, increasing in volume with rising temperature.
Difference between water and other substances:
- Most substances expand as temperature increases, but water first contracts (from 0°C to 4°C) and then expands.
- This unusual property is due to hydrogen bonding in water molecules.
Name of this behaviour: This phenomenon is called anomalous expansion of water.
What is meant by specific heat capacity?
Answer:
Specific heat capacity is the amount of heat energy required to raise the temperature of 1 kg of a substance by 1°C.
It is denoted by the symbol c and is measured in joules per kilogram per degree Celsius (J/kg°C).
Formula:
Q = mcΔT
- Q = Heat energy (Joules)
- m = Mass of the substance (kg)
- c = Specific heat capacity (J/kg°C)
- ΔT = Change in temperature (°C)
Example: The specific heat capacity of water is 4186 J/kg°C, meaning 4186 joules of heat are needed to raise the temperature of 1 kg of water by 1°C.
How will you prove experimentally that different substances have different specific heat capacities?
Scientific and Written Exam Answer:
To prove that different substances have different specific heat capacities, we conduct an experiment using metal spheres made of iron, copper, and lead. These metals are heated uniformly and then tested for heat retention by observing their effect on wax.
Materials Required:
- Three metal spheres (Iron, Copper, Lead) of equal mass
- Wax layer
- Boiling water
- Thread
Procedure:
- Take three metal spheres of equal mass (iron, copper, and lead) and coat their ends with a thin wax layer.
- Heat all three spheres in boiling water for the same duration to ensure uniform temperature.
- Remove them from the boiling water and immediately place them on a surface with their wax-coated ends facing downward.
- Observe the melting rate of the wax in contact with each metal.
Observation:
- The wax under the copper sphere melts faster than the wax under iron.
- The wax under the lead sphere melts the slowest.
Conclusion:
Since all three spheres were heated for the same duration, the difference in wax melting speed indicates that different metals transfer heat at different rates. This confirms that different substances have different specific heat capacities.
Formula for Specific Heat Capacity:
Specific heat capacity \( c \) is given by:
$$ c = \frac{Q}{m\Delta T} $$
Where:
- \( c \) = Specific heat capacity (J/kg·K)
- \( Q \) = Heat energy supplied (Joules)
- \( m \) = Mass of the substance (kg)
- \( \Delta T \) = Change in temperature (K)
Conclusion:
This experiment shows that different metals have different capacities to store heat, proving that substances have different specific heat capacities.
While deciding the unit for heat, which temperature interval is chosen? Why?
Scientific and Written Exam Answer:
The temperature interval chosen while deciding the unit for heat is 1 Kelvin (1 K) or 1 degree Celsius (1°C). This is because the unit of heat is defined based on the amount of heat energy required to raise the temperature of a substance by a specific amount.
Explanation:
- Heat energy is measured in joules (J) in the SI system.
- The specific heat capacity (c) of a substance is defined as the amount of heat required to raise the temperature of 1 kg of the substance by 1 K (or 1°C).
- Thus, the unit of heat is based on a temperature interval of 1 Kelvin (K) or 1°C.
Formula for Heat Energy:
Heat energy \( Q \) is calculated using the formula:
$$ Q = mc\Delta T $$
Where:
- \( Q \) = Heat energy (Joules)
- \( m \) = Mass of the substance (kg)
- \( c \) = Specific heat capacity (J/kg·K)
- \( \Delta T \) = Temperature change (K or °C)
Conclusion:
The temperature interval of 1 K (or 1°C) is chosen because it provides a standard way to measure heat energy and compare different substances easily.
Explain the following temperature vs time graph.
Scientific and Written Exam Answer:
The given graph represents the temperature change of water with respect to time as heat is supplied. It shows different phase transitions of water: from solid (ice) to liquid (water) and then to gas (steam).
Explanation of the Graph:
- Segment AB: Ice is melting at 0°C. The temperature remains constant as the ice changes into liquid water.
- Segment BC: Water is heating up from 0°C to 100°C as heat is supplied, increasing its temperature.
- Segment CD: Water is boiling at 100°C. The temperature remains constant as the liquid changes into steam (gas).
Key Points:
- During phase changes (melting and boiling), the temperature remains constant.
- Heat energy is used for phase transformation instead of increasing temperature.
- Once all the water is converted into steam, further heat increases the temperature of steam.
Scientific Terms:
- Latent Heat of Fusion: The heat required to convert ice into water without changing temperature.
- Latent Heat of Vaporization: The heat required to convert water into steam at constant temperature.
Simple and Understandable Answer:
This graph shows how ice turns into water and then into steam as we heat it.
- First stage (AB): Ice melts into water at 0°C. The temperature does not increase.
- Second stage (BC): Water heats up from 0°C to 100°C.
- Third stage (CD): Water boils at 100°C, changing into steam, and the temperature stays the same.
Example: If we heat an ice cube, it first melts, then the water heats up, and finally, it starts boiling and turning into steam.
Conclusion: This graph helps us understand how substances change from solid to liquid and liquid to gas while maintaining temperature during phase changes.
What is the role of the anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Scientific and Written Exam Answer:
The anomalous behaviour of water refers to its property of expanding when cooled below 4°C, instead of contracting like other liquids. This unique property plays a crucial role in preserving aquatic life in cold regions.
Explanation:
- At temperatures above 4°C, water behaves normally, contracting as it cools.
- However, below 4°C, water expands, making ice less dense than liquid water.
- As a result, ice floats on the surface of water bodies, forming an insulating layer.
- This layer prevents further heat loss and keeps the water beneath from freezing, allowing aquatic life to survive.
Scientific Terms:
- Anomalous Expansion: Water expands instead of contracting when cooled below 4°C.
- Density Inversion: Water at 4°C is denser than water at 0°C, ensuring ice remains on the surface.
Formula:
Density of water decreases as temperature falls below 4°C:
$$ \rho = \frac{m}{V} $$ (where \( \rho \) is density, \( m \) is mass, and \( V \) is volume)
Simple and Understandable Answer:
Water behaves differently from other liquids because it expands when it cools below 4°C. This helps aquatic life survive in cold regions.
- When water freezes, it turns into ice, which is lighter than liquid water.
- The ice stays on the surface, acting like a protective cover.
- The water underneath stays liquid, allowing fish and other aquatic organisms to survive even in extreme cold.
Example: In winter, lakes and ponds freeze only at the top, but fish can still swim in the water below because it doesn’t freeze completely.
Conclusion: The anomalous expansion of water is essential for the survival of aquatic life in cold climates, as it prevents entire water bodies from freezing solid.
How can you relate the formation of water droplets on the outer surface of a bottle taken out of the refrigerator with the formation of dew?
Scientific and Written Exam Answer:
The formation of water droplets on the outer surface of a cold bottle and the formation of dew are both examples of condensation, a physical process where water vapor in the air cools down and changes into liquid form.
Explanation:
- When a bottle is taken out of the refrigerator, its surface is much colder than the surrounding air.
- The warm air around the bottle contains moisture in the form of water vapor.
- As this air comes into contact with the cold surface of the bottle, the temperature of the air drops, causing the water vapor to condense into tiny water droplets.
- A similar process occurs in nature when the Earth's surface cools at night.
- The air near the ground loses heat, and if the temperature drops below the dew point, the moisture in the air condenses into tiny droplets, forming dew.
Scientific Terms:
- Condensation: The process of water vapor turning into liquid due to cooling.
- Dew Point: The temperature at which air becomes saturated, leading to condensation.
Simple and Understandable Answer:
When you take a cold bottle out of the refrigerator, tiny water droplets appear on its surface. This happens because the warm air around the bottle cools down, and the water vapor in the air turns into liquid.
- The same process occurs when dew forms on grass and leaves in the early morning.
- At night, the ground cools down, and the moisture in the air turns into tiny water droplets, forming dew.
Example: Just like how a cold bottle gets wet outside, car windows and grass also get covered with dew in the morning.
Conclusion: Both the water droplets on a bottle and morning dew form due to condensation when warm air meets a cold surface.
In cold regions in winter, the rocks crack due to anomalous expansion of water.
Scientific and Written Exam Answer:
Rocks in cold regions crack in winter due to the anomalous expansion of water. Water exhibits a unique behavior where it expands upon freezing instead of contracting.
Explanation:
- When water enters small cracks or pores in rocks, it remains in liquid form at temperatures above 4°C.
- As the temperature drops below 4°C, water expands instead of contracting, due to its anomalous behavior.
- When water freezes at 0°C, it expands further, exerting high pressure on the rock.
- The repeated freezing and expansion cause the cracks in the rock to widen, eventually breaking the rock into smaller pieces.
Scientific Terms:
- Anomalous Expansion of Water: The unusual property of water where it expands upon freezing instead of contracting.
- Frost Weathering: The process where rock breakdown occurs due to repeated freezing and thawing of water.
Simple and Understandable Answer:
In very cold regions, water can enter small cracks in rocks. When the temperature drops, the water freezes and expands, making the crack bigger. Over time, the rock breaks apart due to this repeated freezing and expansion.
- This happens because water behaves differently from other liquids. Instead of shrinking when frozen, it expands.
- When water freezes inside a rock, it pushes against the rock, causing cracks to grow.
Example: If you put a full bottle of water in the freezer, it can break due to expansion. Similarly, water in rocks causes them to crack in winter.
Conclusion: The expansion of freezing water puts pressure on rocks, leading to cracks and breaking, especially in extremely cold places.
Answer the following:
What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Solution 1: Scientific and Written Exam Answer
The heat energy required to change a substance from one state to another at a constant temperature is called the latent heat of a substance.
When latent heat is released from a substance, the intermolecular forces become stronger, leading to a phase transition:
- When a liquid releases latent heat of fusion, the stronger molecular bonds pull the molecules closer, and the substance transitions into a solid state (Freezing/Solidification).
- When a gas releases latent heat of vaporization, the molecules lose energy, bond more tightly, and the substance transforms into a liquid state (Condensation).
Formula for Latent Heat:
$$ Q = mL $$
Where,
- Q = Heat energy absorbed or released (Joules)
- m = Mass of the substance (kg)
- L = Specific latent heat (J/kg)
This process happens without a change in temperature but results in a phase change due to energy transfer.
Solution 2: Simple and Understandable Answer
Latent heat is the hidden heat energy that helps a substance change its state without changing its temperature. For example, when ice melts into water, it absorbs heat but stays at 0°C until all the ice has melted.
When heat is released:
- Water turns into ice (Freezing), just like how water in a freezer becomes ice cubes.
- Steam turns back into water (Condensation), like how water droplets form on a cold surface.
This happens because the released heat goes into the surroundings, making the molecules come closer and changing the substance's form.
Which principle is used to measure the specific heat capacity of a substance?
Solution 1: Scientific and Written Exam Answer
The principle used to measure the specific heat capacity of a substance is the Principle of Calorimetry. According to this principle:
"When two bodies at different temperatures are brought into thermal contact, heat lost by the hotter body is equal to the heat gained by the colder body, provided no heat is lost to the surroundings."
Mathematically, this is expressed as:
$$ m_1 c_1 \Delta T_1 = m_2 c_2 \Delta T_2 $$
Where,
- m₁, m₂ = Mass of the two substances (kg)
- c₁, c₂ = Specific heat capacities of the substances (J/kg·K)
- ΔT₁, ΔT₂ = Change in temperature of the substances (K or °C)
This principle is applied in a calorimeter to determine the specific heat capacity of an unknown substance.
Solution 2: Simple and Understandable Answer
The Principle of Calorimetry is used to measure how much heat energy a substance can store. It states that:
"Heat lost by a hot object is equal to the heat gained by a cold object."
For example, if you mix hot water and cold water in a cup, the hot water cools down, and the cold water warms up. The total heat remains the same.
This principle helps scientists measure the specific heat capacity of different materials using a device called a calorimeter.
Explain the role of latent heat in the change of state of a substance.
Solution 1: Scientific and Written Exam Answer
Latent heat is the amount of heat energy required to change the state of a substance without changing its temperature.
During a phase change, latent heat is either absorbed or released:
- Latent Heat of Fusion (Melting): The heat energy required to change a solid into a liquid at its melting point without temperature change.
- Latent Heat of Vaporization (Boiling): The heat energy required to change a liquid into a gas at its boiling point without temperature change.
Mathematically, latent heat is given by:
$$ Q = mL $$
Where:
- Q = Heat energy absorbed or released (Joules)
- m = Mass of the substance (kg)
- L = Latent heat (J/kg)
When a substance absorbs latent heat, its molecular bonds break, allowing a change of state (e.g., ice melting into water). When latent heat is released, molecular bonds strengthen, leading to a change of state (e.g., water freezing into ice).
Solution 2: Simple and Understandable Answer
Latent heat is the hidden heat that helps a substance change its state without increasing its temperature.
For example:
- When ice melts into water, it absorbs heat but remains at 0°C until fully melted.
- When water boils into steam, it absorbs heat but stays at 100°C until completely converted into gas.
This means that heat energy is used to break the bonds between molecules instead of increasing temperature.
Similarly, when steam condenses into water or water freezes into ice, it releases latent heat to the surroundings, helping in the phase change.
On what basis and how will you determine whether air is saturated with vapour or not?
Solution 1: Scientific and Written Exam Answer
Air is considered saturated with vapour when it contains the maximum amount of water vapour it can hold at a given temperature and pressure. This means that the air has reached its maximum capacity to hold moisture, and any additional water vapour will condense into liquid form.
Basis for Determining Saturation:
- Relative Humidity (RH): If the relative humidity reaches 100%, the air is saturated with water vapour.
- Dew Point Temperature: The temperature at which air becomes saturated, and condensation begins. If the actual temperature is equal to the dew point temperature, the air is saturated.
- Condensation and Fog Formation: If visible condensation (such as mist, fog, or dew) forms, it indicates that the air has reached saturation.
Determining Whether Air is Saturated:
We can determine air saturation using the formula:
$$ RH = \left( \frac{e}{E} \right) \times 100 $$
Where:
- RH = Relative Humidity (%)
- e = Actual vapour pressure (Pa)
- E = Saturation vapour pressure at the given temperature (Pa)
If RH = 100%, the air is saturated. If RH is less than 100%, the air can still absorb more moisture.
Another way to determine saturation is by cooling air gradually. If moisture starts condensing at a particular temperature (dew point), it confirms that the air is saturated.
Solution 2: Simple and Understandable Answer
We say that air is saturated when it can no longer hold more water vapour. At this point, the water in the air starts turning into tiny droplets, forming clouds, fog, or dew.
How to Check if Air is Saturated:
- Look for Fog or Dew: If water droplets form on surfaces or in the air, the air is saturated.
- Check the Relative Humidity: If a weather report says the humidity is 100%, the air is fully saturated.
- Cooling Test: If air cools down and dew forms, it means the air has reached its limit and is saturated.
Example: On cold winter mornings, you often see fog or water droplets on windows. This happens because the air is fully saturated, and the extra water vapour turns into tiny water droplets.
Read the following paragraph and answer the questions.
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to the gain of energy, and the temperature of the hot object goes on decreasing due to the loss of energy. The change in temperature continues till the temperatures of both objects attain the same value. In this process, the cold object gains heat energy, and the hot object loses heat energy. If the system of both objects is isolated from the environment by keeping it inside a heat-resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
Questions and Answers:
Solution 1: Scientific and Written Exam Answer
i. Heat is transferred from where to where?
Heat is transferred from the hot object to the cold object due to the temperature difference between them. This transfer continues until both objects reach thermal equilibrium.
ii. Which principle do we learn about from this process?
The process demonstrates the Principle of Conservation of Energy, specifically applied to the transfer of heat energy.
iii. How will you state the principle briefly?
The principle states that "In an isolated system, the total heat lost by a hot body is equal to the total heat gained by a cold body, assuming no external heat exchange occurs."
Mathematically, it can be represented as:
$$ Q_{\text{lost}} = Q_{\text{gained}} $$
iv. Which property of the substance is measured using this principle?
The property measured using this principle is the specific heat capacity (c) of a substance. This property determines the amount of heat energy required to raise the temperature of a unit mass of the substance by 1°C.
Solve the following problem:
Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3°C and B by 5°C. Which object has more specific heat? And by what factor?
Solution 1: Scientific and Written Exam Answer
We know that the specific heat capacity (c) is given by the formula:
$$ Q = mc\Delta T $$
Since the heat supplied (Q) and mass (m) are the same for both objects, we can compare their specific heat capacities using:
$$ c_A \times \Delta T_A = c_B \times \Delta T_B $$
Rearranging for the ratio:
$$ \frac{c_A}{c_B} = \frac{\Delta T_B}{\Delta T_A} $$
Given,
Mass of A = Mass of B = 1 g
Temperature rise in A, ΔTA = 3°C
Temperature rise in B, ΔTB = 5°C
Substituting the values:
$$ \frac{c_A}{c_B} = \frac{5}{3} $$
Thus, the specific heat capacity of object A is 5/3 times that of object B.
Solution 2: Simple and Understandable Answer
Specific heat capacity (c) tells us how much heat is needed to change the temperature of a substance. The formula is:
$$ Q = mc\Delta T $$
Since both objects received the same heat and have the same mass, the object with the smaller temperature rise must have the higher specific heat.
Object A's temperature increased by 3°C, while object B's temperature increased by 5°C. Since A’s temperature changed less, it has a higher specific heat.
By comparing their values:
$$ \frac{c_A}{c_B} = \frac{5}{3} $$
So, object A has a specific heat that is 1.67 times higher than that of object B.
Example: Water heats up more slowly than metal because water has a higher specific heat capacity.
Solve the following problem:
Liquid ammonia is used in ice factories for making ice from water. If water at 20°C is to be converted into 2 kg ice at 0°C, how many grams of ammonia are to be evaporated?
Given:
- Mass of water = 2 kg = 2000 g
- Initial temperature of water (Ti) = 20°C
- Final temperature of ice (Tf) = 0°C
- Specific heat capacity of water (c) = 1 cal/g°C
- Latent heat of fusion of ice (Lf) = 80 cal/g
- Latent heat of vaporization of ammonia = 341 cal/g
Solution 1: Scientific and Written Exam Answer
To convert water at 20°C into ice at 0°C, we need to:
- Cool the water from 20°C to 0°C.
- Freeze the water into ice.
The total heat (Q) to be removed consists of two parts:
1. Heat required to cool water from 20°C to 0°C:
$$ Q_1 = mc\Delta T $$
Substituting values:
$$ Q_1 = (2000)(1)(20) $$
$$ Q_1 = 40000 \text{ cal} $$
2. Heat required to convert water at 0°C into ice at 0°C:
$$ Q_2 = mL_f $$
Substituting values:
$$ Q_2 = (2000)(80) $$
$$ Q_2 = 160000 \text{ cal} $$
Total heat to be removed:
$$ Q = Q_1 + Q_2 $$
$$ Q = 40000 + 160000 $$
$$ Q = 200000 \text{ cal} $$
Now, we calculate the mass of ammonia (mNH3) required to remove this heat. Since the latent heat of vaporization of ammonia is 341 cal/g, we use:
$$ Q = m_{NH3} \times L_{NH3} $$
Rearranging:
$$ m_{NH3} = \frac{Q}{L_{NH3}} $$
Substituting values:
$$ m_{NH3} = \frac{200000}{341} $$
$$ m_{NH3} \approx 586.5 \text{ g} $$
Answer: Approximately 586.5 g of ammonia must be evaporated.
Solution 2: Simple and Understandable Answer
To make ice from water at 20°C, we must first cool the water down to 0°C and then freeze it into ice. This process requires removing heat.
✔ The heat needed to cool 2 kg of water from 20°C to 0°C is:
$$ Q_1 = 40000 \text{ cal} $$
✔ The heat needed to freeze 2 kg of water into ice is:
$$ Q_2 = 160000 \text{ cal} $$
✔ Total heat to be removed:
$$ Q = 200000 \text{ cal} $$
✔ Ammonia removes heat by evaporating. Since each gram of ammonia removes 341 cal, we calculate:
$$ m_{NH3} = \frac{200000}{341} \approx 586.5 \text{ g} $$
So, about 586.5 g of ammonia must evaporate to make the ice.
Example: Just like ice melts in a glass of water by absorbing heat, ammonia helps make ice by taking away heat when it evaporates.
Solve the following problem:
A thermally insulated pot has 150 g ice at temperature 0°C. How much steam of 100°C has to be mixed to it, so that water of temperature 50°C will be obtained?
Given:
- Mass of ice (mice) = 150 g
- Initial temperature of ice = 0°C
- Final temperature of water = 50°C
- Latent heat of fusion of ice (Lf) = 80 cal/g
- Latent heat of vaporization of water (Lv) = 540 cal/g
- Specific heat capacity of water (c) = 1 cal/g°C
- Temperature of steam = 100°C
Solution with Step-by-Step Explanation
Step 1: Understanding the Heat Exchange
✔ The system is thermally insulated, meaning no heat is lost to the surroundings.
✔ Heat is transferred from the steam (which condenses and cools) to the ice (which melts and then heats up).
Step 2: Heat Required to Convert Ice at 0°C into Water at 50°C
To convert ice at 0°C into water at 50°C, three processes take place:
- 1️⃣ Ice melts into water at 0°C.
- 2️⃣ Water at 0°C heats up to 50°C.
Heat required to melt the ice:
$$ Q_1 = m_{ice} \times L_f $$
Substituting values:
$$ Q_1 = 150 \times 80 $$
$$ Q_1 = 12000 \text{ cal} $$
Heat required to heat the melted ice from 0°C to 50°C:
$$ Q_2 = m_{ice} \times c \times \Delta T $$
Substituting values:
$$ Q_2 = 150 \times 1 \times (50 - 0) $$
$$ Q_2 = 150 \times 50 $$
$$ Q_2 = 7500 \text{ cal} $$
Total heat required:
$$ Q_{total} = Q_1 + Q_2 $$
$$ Q_{total} = 12000 + 7500 $$
$$ Q_{total} = 19500 \text{ cal} $$
Step 3: Heat Given by Steam to Convert into Water at 50°C
Steam at 100°C first condenses into water at 100°C, then cools down to 50°C.
- 1️⃣ Steam condenses into water at 100°C.
- 2️⃣ Water at 100°C cools down to 50°C.
Heat released when steam condenses into water:
$$ Q_3 = m_{steam} \times L_v $$
Substituting values:
$$ Q_3 = m_{steam} \times 540 $$
Heat released when condensed water cools from 100°C to 50°C:
$$ Q_4 = m_{steam} \times c \times \Delta T $$
Substituting values:
$$ Q_4 = m_{steam} \times 1 \times (100 - 50) $$
$$ Q_4 = m_{steam} \times 50 $$
Total heat released:
$$ Q_{released} = Q_3 + Q_4 $$
$$ Q_{released} = m_{steam} \times 540 + m_{steam} \times 50 $$
$$ Q_{released} = m_{steam} \times (540 + 50) $$
$$ Q_{released} = m_{steam} \times 590 $$
Step 4: Equating Heat Gained and Heat Lost
Since the heat gained by ice must be equal to the heat lost by steam:
$$ 19500 = m_{steam} \times 590 $$
Solving for msteam:
$$ m_{steam} = \frac{19500}{590} $$
$$ m_{steam} \approx 33.05 \text{ g} $$
Final Answer:
The amount of steam required is approximately 33.05 g.
Solution Summary:
- We calculated the heat required to melt ice and heat it to 50°C.
- We calculated the heat released when steam condenses and cools.
- We equated both to find that 33.05 g of steam is required.
Example: Just like pouring hot tea into a glass of ice water changes the temperature, steam mixes with ice to form warm water.
Solve the following problem:
A calorimeter has mass 100 g and specific heat 0.1 kcal/kg°C. It contains 250 g of liquid at 30°C, having a specific heat of 0.4 kcal/kg°C. If we drop a piece of ice of mass 10 g at 0°C, what will be the final temperature of the mixture?
Given:
- Mass of the calorimeter (mc) = 100 g = 0.1 kg
- Specific heat of calorimeter (cc) = 0.1 kcal/kg°C
- Mass of liquid (ml) = 250 g = 0.25 kg
- Specific heat of liquid (cl) = 0.4 kcal/kg°C
- Initial temperature of the liquid and calorimeter (Tinitial) = 30°C
- Mass of ice (mice) = 10 g = 0.01 kg
- Latent heat of fusion of ice (Lf) = 80 kcal/kg
- Specific heat of water (cw) = 1 kcal/kg°C
- Final temperature = T
Solution with Step-by-Step Explanation
Step 1: Understanding the Heat Exchange
✔ The system is thermally insulated, meaning no heat is lost to the surroundings.
✔ Heat is transferred from the warm liquid and calorimeter to the ice, which melts and then heats up.
Step 2: Heat Lost by the Liquid and Calorimeter
Both the liquid and calorimeter will lose heat as they cool down to the final temperature (T).
Heat lost by the liquid:
$$ Q_l = m_l \times c_l \times (T_{\text{initial}} - T) $$
$$ Q_l = 0.25 \times 0.4 \times (30 - T) $$
$$ Q_l = 0.1 \times (30 - T) $$
Heat lost by the calorimeter:
$$ Q_c = m_c \times c_c \times (T_{\text{initial}} - T) $$
$$ Q_c = 0.1 \times 0.1 \times (30 - T) $$
$$ Q_c = 0.01 \times (30 - T) $$
Total heat lost:
$$ Q_{\text{lost}} = Q_l + Q_c $$
$$ Q_{\text{lost}} = 0.1 (30 - T) + 0.01 (30 - T) $$
$$ Q_{\text{lost}} = 0.11 (30 - T) $$
Step 3: Heat Gained by Ice
✔ Ice first melts into water at 0°C.
✔ The melted water then heats up to the final temperature T.
Heat required to melt the ice:
$$ Q_{\text{melt}} = m_{\text{ice}} \times L_f $$
$$ Q_{\text{melt}} = 0.01 \times 80 $$
$$ Q_{\text{melt}} = 0.8 \text{ kcal} $$
Heat required to raise the melted ice temperature to T:
$$ Q_{\text{heat}} = m_{\text{ice}} \times c_w \times (T - 0) $$
$$ Q_{\text{heat}} = 0.01 \times 1 \times (T - 0) $$
$$ Q_{\text{heat}} = 0.01 T $$
Total heat gained by ice:
$$ Q_{\text{gained}} = Q_{\text{melt}} + Q_{\text{heat}} $$
$$ Q_{\text{gained}} = 0.8 + 0.01T $$
Step 4: Equating Heat Lost and Heat Gained
Since no heat is lost to the surroundings:
$$ Q_{\text{lost}} = Q_{\text{gained}} $$
$$ 0.11 (30 - T) = 0.8 + 0.01T $$
Step 5: Solving for T
Expanding:
$$ 3.3 - 0.11T = 0.8 + 0.01T $$
Rearrange:
$$ 3.3 - 0.8 = 0.11T + 0.01T $$
$$ 2.5 = 0.12T $$
$$ T = \frac{2.5}{0.12} $$
$$ T \approx 20.83°C $$
Final Answer:
The final temperature of the mixture is approximately 20.83°C.
Solution Summary:
- We calculated the heat lost by the liquid and calorimeter.
- We calculated the heat gained by the ice as it melts and heats up.
- We equated both and solved for the final temperature.
- The final temperature of the mixture is 20.83°C.
Example: Just like adding an ice cube to a cup of warm water, the ice melts and the overall temperature drops.