Question

A 50·g ice cube (at 0°C) is placed in an insulated cup with 250·g of water which is at 46°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 heat will the ice have to absorb from the water to completely melt (and turn into 0°C water)? ___ cal.

(b) Find the temperature of the 250·g of water initially at 46°C after it loses the heat required to melt the ice.? ___ °C.

(c) How much heat will the 250·g of warmer water have to transfer to the cooler water in order for its temperature to drop by 1°C? ____ cal.

(d) How much will the temperature of the 50·g of cooler water rise when it gains the heat lost by the warmer water as it cools by 1°C? ____°C.

(e) 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? _____°C.

Answer 1

PART A:

The latent heat of fusion for ice is

The mass of the ice cube is

So, the heat absorbed by the ice to completely melt is

PART B:

The specific heat of the water is

The mass of the water is

The heat lost by the water is

Which gives us the difference in the temperature is

So, the final temperature of the water is

PART C:

We have

So, the heat required to drop the temp by 1 deg is

PART D:

The heat lost by warmer water is 250 cal.

The mass of the cooler water is 50 g.

So, the rise in temperature is

PART E:

Let us assume the equilibrium temperature is T

The heat lost by the warmer water must be equal to heat gain by colder water. i.e

So, the equilibrium temperature is 25 ^oC.