Question

A 100 g copper bowl contains 500 g of water, both at 20◦C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5 g of the water being converted to steam, and the final temperature of the system is 100◦C. What is the original temperature of the cylinder?

Answer 1

Solution:

Since the water boils, with 5 g being converted into steam, we can find the energy gained by water by adding the energy required to take all 500 g of the water from 20◦C to 100◦C, and the energy required to convert 5 g of water to steam at 100◦C.

Hence;

Energy gained by water = 0.5 * 4186 * 80 + 0.005 *  22.6 × 10⁵ = 178740 J

Since the bowl remains solid throughout, the energy transferred to it will be;

Energy gained by bowl = (0.100 kg) C_CU (100 − 20) = 3080 J

now as we know;

The total energy lost by the cylinder is equal to the sum of the energy gained by the water and the bowl

=> 0.3 * 385 * (T-100) = 3080 + 178740

=> T = 1674.2 ^oC