A 60 ·g ice cube (at 0 °C) is placed in an insulated cup with 480 ·g of water wh

Question

A 60 ·g ice cube (at 0 °C) is placed in an insulated cup with 480 ·g of water which is at 55 °C. The latent heat of fusion for ice is 80 ·cal/g and the specific heat of water is 1.0 ·cal/g/ °C.

A) How much will the temperature of the 60 ·g of cooler water rise when it gains the heat lost by the warmer water as it cools by 1 °C? answer in °C.

B) Using your results from the two previous parts, continue to follow the process of heat transfer between the warmer and cooler water until they come to equilibrium. At what temperature will they reach thermal equilibrium? answer in
°C.

Explanation / Answer

(A) Heat lost = heat gained..... 60*1*dt = 480*1*1..... dt = 8 C..... so, if warmer water cools by 1 C, cooler water heats by 8 C.... (B) Let equilibrium temperature = t.... Heat lost = heat gained..... 60*80 + 60*1*t = 480*(55-t) 540t = 21600 t = 40 degrees

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