A 100 g copper bowl contains 240 g of water, both at 20.0°C. A very hot 390 g co

Question

A 100 g copper bowl contains 240 g of water, both at 20.0°C. A very hot 390 g copper cylinder is dropped into the water, causing the water to boil, with 22.0 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature (in Celsius) 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

(a) Energy transferred to water as heat is Ew = 240*1*(100-20) + (22/1000)*539 = 19211.858 cal (b) Eb = 100*0.0923*(100-20) = 738.4 cal (c) Heat lost by the hot body = Heat gained by the cold body 390*0.0923*(T-100) = 19211.858+738.4 = 19950.258 T = 654.22 degrees C

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