WBCHSE Class 11 Physics For Calorimetry Questions and Answers

Calorimetry Long Answer Type Questions And Answers

Question 1. Equal masses of milk and water are taken in two identical kettles and are heated by the same source. The rate of rise of the temperature of milk is found to be higher than that of water. Explain.
Answer:

Since both the kettles are heated by the same source, in the same interval of time amount of heat (H) absorbed is the same. Now if the rate of absorption of heat \(\left(\frac{H}{\text { time }}\right)\) and mass (m) are fixed, then from the equation H = mst we get,

st ∝ constant

∴ \(t \propto \frac{1}{s}\)

Now, since specific heat of milk is less than that of water, rate of increase in temperature of the milk is greater than that of water of the same mass.

Question 2. Milk and water of equal mass are taken in two similar vessels at room temperature. Both of them are to be heated from room temperature to a certain higher temperature. Which will take less heat and why?
Answer:

Let mass of milk = m = mass of water

Increase in temperature of milk = t = increase in temperature of water

Let specific heats of milk and water be s_m and s_w.

Heat absorbed by milk, H_m = ms_mt

Heat absorbed by water, H_w = ms_wt

∴ \(s_m<s_w, \quad H_m<H_w\)

So, less heat is required to warm milk.

Question 3. 1 kg of iron at 100°C melts more Ice than 1 kg of lead at 100°C. Explain why.
Answer:

Specific heat of iron is more than that of lead. Hence, for the same fall in temperature, iron supplies more heat to the ice than lead of the same mass does. Hence, iron melts relatively more ice.

Question 4. What is the advantage of taking water in hot water bottles?
Answer:

Water has a specific heat higher than everything but ammonia. Hence, a fixed mass of water gains more heat than any other liquid of the same mass for the same rise in temperature.

Consequently, it loses more heat during cooling. This heat is used for fomentation. So a ot water bottle remains effective for a long period of time.

Question 5. Two copper spheres of the some external radius, one solid but the other hollow, are heated up to a certain temperature and then are allowed to cool under similar conditions. Which sphere will cool faster?
Answer:

The hollow sphere will cool faster.

As both are made of copper, the specific heat s is the same.

Now, H = mst, or t = H/ms, where t = rate of fall of temperature. As the mass of the hollow sphere is less, t will be higher for it. So it will cool faster.

Question 6. State whether the fundamental law of calorimetry is applicable in the following cases:

1. Sugar is added to water taken in a calorimeter,
2. A chemical reaction occurs between a solid and a liquid in a calorimeter,
3. Calorimeter is kept open in air.

Answer:

While dissolving in water, sugar will absorb the necessary heat of the solution. If it is not taken into consideration, the fundamental law of calorimetry cannot be applied.

• During a chemical reaction, heat is evolved or absorbed. If this heat is not taken into consideration, the fundamental law of calorimetry cannot be applied.
• If the calorimeter is kept open in air, heat exchange occurs with the surroundings. Hence, the fundamental law of calorimetry will not be applicable.

Question 7. Two small spherical balls of the same mass, made of copper and of lead are heated up to the same tern- perature and are placed on a thick wax slab. What will happen and what can we conclude from it?
Answer:

The copper ball sinks in the wax slab more than the lead ball does, because the rate of melting of wax due to the copper ball is greater. From this we conclude that the heat loss of the copper ball is greater than that of the lead ball. But, the mass and the decrease in temperatures of both the balls are equal. Therefore, from the equation H = mst we conclude, the specific heat (s) of copper is higher.

Question 8. If two bodies of equal mass but of different materials are supplied equal amounts of heat, which one will have higher rise in temperature?
Answer:

Given, Heat H  = mst i.e,  temperature rise, t = H/ms

Since for the two bodies, heat supplied H and mass m are equal, \(t \propto \frac{1}{s}\)

Hence, the body with comparatively less specific heat will have a higher rise in temperature.

Question 9. 100 g of water and 100 g of iron are heated up to the same temperature and are separately added to 50 g of water at a lower temperature, kept in two identical vessels. Compare the temperature changes in the two vessels.
Answer:

Temperatures of both the vessels will increase. The specific heat of iron is much less than that of water (s_water = 1 and s_iron = 0.117). The initial temperatures of the given specimens of 100 g of water and 100 g of iron are equal.

But, due to its higher specific heat, water will lose more heat than iron will. Hence, the final temperature of the vessel in which 100 g of hot water is added will be higher.

Question 10. When a hot body heats up a cold body, is the temperature change of both the bodies the same? Explain the answer.
Answer:

The change in temperature of the two bodies would differ. Heat lost by the hot body or gained by the cold body is equal to the product of

1. Mass
2. Specific heat and
3. Change in temperature.

Hence, the body having higher mass and higher specific heat will have less change in temperature for the same amount of heat transfer.

Question 11. Why calorimeters are made of metal (mostly copper) instead of glass?
Answer:

Due to higher conductivity of metals, in a metallic calorimeter thermal equilibrium is achieved very fast. Also specific heat of metals is lower than that of glass.

• That is why the water equivalent of a metallic calorimeter is lower than that of a glass calorimeter of the same mass. Therefore, for the same rise in temperature heat gained by metallic calorimeter is less than that gained by glass calorimeter.
• As a result the final temperature of the mixture will be higher and will result less error in measurement.
• Copper has the highest conductivity (other than silver) among all the metals, and copper is much cheaper than silver, that is why copper is mostly used as the material of a calorimeter.

Question 12. The temperature of a furnace is more than 500°C. Discuss a method of measuring the temperature of the furnace with the help of a thermometer graduated up to 100°C.
Answer:

Let the temperature of the furnace = t°C. A small metal piece of mass m and specific heat s is put into the furnace. Obviously, the melting point of the metal should be greater than the temperature of the furnace so that it does not melt.

Now, water of mass M is taken in a vessel of water equivalent W and its temperature is recorded. Let the temperature be  t₁(< 100°C).

Now the hot metal piece is dropped in that water. As a result, a small quantity of water is vapourised which is negligible. After a while, when the metal piece and water reach thermal equilibrium, the final temperature is measured by the thermometer.

If that temperature is t₂°C, heat lost by the metal piece = ms(t- t₂) and heat gained by the vessel and water = (W+ M)s_w(t₂ – t₁), where s_w is the specific heat of water.

From calorimetric principle, ms(t-t₂) = (W+ M)s_w(t₂ – t₁)

Value of l can be calculated from this equation if values of all the other quantities are known. Thus, measuring temperatures t₁ and t₂ with the help of a thermometer graduated up to 100°C, a much greater temperature t can he determined.

Question 13. The diameter of an iron sphere and the length of the side of an iron cube are equal. Initially they are at the same temperature. If they take the same amount of heat, whose final temperature will be higher?
Answer:

If x be the diameter of the sphere, its volume is, \(V_1=\frac{4}{3} \pi\left(\frac{x}{2}\right)^3\)

The length of the side of the cube is x.

The volume of the cube is, V₂ = x³

∴ \(\frac{V_1}{V_2}=\frac{\frac{4}{3} \pi \frac{x^3}{8}}{x^3}=\frac{\pi}{6}<1 \quad therefore V_1<V_2\)

Since the volume of the cube is greater than that of the sphere, the mass of the cube is also greater. So, if an equal amount of heat is taken by the cube, its rise in temperature will be smaller, i.e., the rise in temperature of the sphere will be higher.