A 180 g copper bowl contains 190 g of water, both at 19.0°C. A very hot 450 g co

Question

A 180 g copper bowl contains 190 g of water, both at 19.0°C. A very hot 450 g copper cylinder is dropped into the water, causing the water to boil, with 15.2 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 in water = 190*1*(100-19) = 16910.0 Cal

heat in steam = 15.2*539 = 8192.8 Cal

So heat transferred to water = 16910.0+8192.8 = 25102.8 Cal

Heat transferred to bowl = 180*0.0923*(100-19) = 1478.646 Cal

Total heat trannsferred =  25102.8 Cal + 1478.646 = 26581.446 Cal

So temp of Block = 26581.446/0.0923*450 +100 = 739.977 C

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