A 140 g copper bowl contains 170 g of water, both at 19.0°C. A very hot 350 g co

Question

A 140 g copper bowl contains 170 g of water, both at 19.0°C. A very hot 350 g copper cylinder is dropped into the water, causing the water to boil, with 16.1 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.

A 140 g copper bowl contains 170 g of water, both at 19.0°C. A very hot 350 g copper cylinder is dropped into the water, causing the water to boil, with 16.1 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.

A 140 g copper bowl contains 170 g of water, both at 19.0°C. A very hot 350 g copper cylinder is dropped into the water, causing the water to boil, with 16.1 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.

Explanation / Answer

Heat gained by bowl = 140 *0.0923* [100 - 19] = 1046.68 cal.
Heat gained by 170 gm of water = 170 *1*[81] =13770 cal.
Heat used to convert 16.1 g of water into steam = mL
= 539*16.1 = 8677.9 cal.
Total calories gained by water and bowl= 23494.58 cal.
Heat lost by 350g of copper is 350* 0.0923*[ - 100]
= 32.305[ - 100] cal.
Equating the two
32.305[ - 100] = 23494.58
[ - 100] = 727.27
[] = 827.27° C.

a) 13770 + 8677.9 = 22447.9 cal.
b) 1046.68 cal
c) 827.27 ° C.

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