10 g of ice at -10°C is mixed with 10 g of water at 0°C. The amount of heat required to raise the temperature of mixture to 10°C is
- (a)400 cal
- (b)550 cal
- (c)1050 cal
- (d)1200 cal
Correct - C, 1050 cal. Take the specific heat of ice as 0.5 cal per gram per degree Celsius, the latent heat of fusion of ice as 80 cal per gram, and the specific heat of water as 1 cal per gram per degree Celsius. First, warming 10 g of ice from -10 C to 0 C needs 10 * 0.5 * 10 = 50 cal. Next, melting that 10 g of ice at 0 C needs 10 * 80 = 800 cal. Finally, the 10 g of melt-water together with the original 10 g of water - 20 g in all - is warmed from 0 C to 10 C, needing 20 * 1 * 10 = 200 cal. The total heat is 50 + 800 + 200 = 1050 cal.
- (a)400 cal — 400 cal is far too small - it leaves out the large latent heat (800 cal) needed to melt the ice at 0 C.
- (b)550 cal — 550 cal counts some sensible heating but omits or under-counts the 800 cal latent heat of fusion, so it falls well short of the true total.
- (d)1200 cal — 1200 cal overshoots - it adds heat that is not required, for instance by warming or melting a larger mass or using a wrong specific heat.
Taking a substance through a phase change needs two kinds of heat: sensible heat (mass times specific heat times temperature change) to change the temperature, and latent heat (mass times latent heat) to change the state at constant temperature. To turn cold ice into warm water you must warm the ice, melt it (a large latent heat of fusion), and then warm all the resulting water.
Break the process into stages and never skip the latent-heat step at 0 C - melting dominates the total, 800 of the 1050 cal. Remember to warm the combined 20 g of water in the last stage, not just 10 g.
- The specific heat of ice is about 0.5 cal per gram per degree Celsius, and of water is 1 cal per gram per degree Celsius.
- The latent heat of fusion of ice is about 80 cal per gram.
- Sensible heat is mass times specific heat times temperature change; latent heat is mass times latent heat at constant temperature.
- Here the stages give 50 cal (warm the ice), 800 cal (melt it) and 200 cal (warm 20 g of water), totalling 1050 cal.
- Skipping the latent heat of fusion at 0 C, which is the largest single term.
- Warming only 10 g in the final step instead of the full 20 g of water.
Asked as a calorimetry sum mixing ice and water and asking for heat to reach a final temperature, or for the final state.
Consider the following statements: 1. Steam at 100 C and boiling water at 100 C contain the same amount of heat. 2. Latent heat of fusion of ice is equal to the latent heat of vaporization of water. 3. In an air-conditioner, heat is extracted from the room air at the evaporator coils and is rejected out at the condenser coils. Which of these statements is/are correct?
- (a) 1 and 2
- (b) 2 and 3
- (c) Only 2
- (d) Only 3
Answer(d) Only 3
Same latent-heat family. That question turns on latent heat of fusion and vaporization (steam carries extra latent heat beyond boiling water) — the very concept that makes melting the ice the dominant term in this calorimetry sum.
The quantity of heat needed to change unit mass of a substance from liquid to vapour without changing the temperature, is called
- (a) specific heat
- (b) specific latent heat
- (c) thermal capacity
- (d) heat energy
Answer(b) specific latent heat
Defines latent heat — the constant-temperature heat of a phase change. It is the same latent-heat term (of fusion) that makes melting the ice the biggest slice of the total heat in this calorimetry problem.
- practice — not a real PYQ
The heat needed to melt 10 g of ice at 0 C into water at 0 C (latent heat of fusion 80 cal per gram) is
- (a)80 cal
- (b)800 cal
- (c)8 cal
- (d)8000 cal
Answer(b) 800 cal - that is 10 g times 80 cal per gram.
- practice — not a real PYQ
The heat required to raise 20 g of water from 0 C to 10 C (specific heat 1 cal per gram per degree Celsius) is
- (a)20 cal
- (b)200 cal
- (c)2000 cal
- (d)100 cal
Answer(b) 200 cal - that is 20 * 1 * 10.