A 160 g copper bowl contains 180 g of water, both at 19.0°C. A very hot 430 g co

Question

A 160 g copper bowl contains 180 g of water, both at 19.0°C. A very hot 430 g copper cylinder is dropped into the water, causing the water to boil, with 12.7 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

Use the formula mct. Heat gained by bowl = 100 *0.0923* [100 - 19] = 747.63 cal. Heat gained by 180 gm of water = 180 *1*[81] =14580 cal. Heat used to convert 3.5 g of water into steam = mL = 539*3.5 = 1886.5 cal. Total calories gained by water and bowl= 17214.13 cal. --------------------------------------… Heat lost by 360g of copper is 360* 0.0923*[? - 100] = 33.228[? - 100] cal. ============================== Equating the two 33.228[? - 100] = 17214.13 [?] = 618.06° C. --------------------------------------… a) 14580 + 1886.5 = 16466.5 cal. b) 747.63 cal c) 618.06° C.

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