Physics 12.77 FINAL question, ice cup placed in aluminum cup, what is the mass of cup??

Question

A 170kg ice cube at -10 degrees Celsius is placed in an aluminum cup whose initial temperature is 70 degrees Celsius. The system comes to an equilibrium temperature of 20 degrees Celsius. What is the mass of the cup??

I tried a method on here earlier, and i got the wrong answer, can someone help get the correct way of doing it please.

Answer 1

specific heat of ice is 2.06 kJ/kgC

heat of fusion of ice is 334 kJ/kg

specific heat of water is 4.186 kJ/kgC

specific heat of Aluminum is 0.900 kJ/kgC

note that 170 kg is a very large ice cube! perhaps it should be 170g, in which case, divide the answer by 1000.

There are 4 stages to the problem.

1. the ice warms from -10 to 0, providing (-) energy to cool the Al

2. ice melts, providing (-) energy to cool the Al

3. water warms from 0 to 20, providing (-) energy to cool the Al

4. Al cools from 70 to 20, giving it's energy up

1. E = 2.06 kJ/kgC x 170kg x 10C = 3502 kJ

2. E = 334 kJ/kg x 170 kg = 56780 kJ

3. E = 4.186 kJ/kgC x 170 x 20C = 14232 kJ

4  E = 0.900 kJ/kgC x M x (70C-20C) = 45M kJ

Set the sum of the first 3 equal to the last one and solve for M

3502 kJ + 56780 kJ + 14232 kJ = 45M kJ

45M = 74514

M = 1656 kg

please check the arithmetic.

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