A 130 g copper bowl comtains 130 g of water, both at 19.0 (degrees C). A very ho

Question

A 130 g copper bowl comtains 130 g of water, both at 19.0 (degrees C). A very hot 410 g copper cylinder is dropped into the water, causing the water to boil, with 8.70 g being converted to steam. the final temperature if the system is 100(degrees C). Neglect energy transfer with the enviroment. (a) how much energy (in calories) is transfrred to the water as heat? (b) how much to the bowl? (c) what is the origonal temperature (in celsius) of the cylinder? The specific heat of water is 1 cal/g. K, and of copper is .0923 cal/g.K. The latent heat of vaporization of water is 539 cal/kg.

Explanation / Answer

latent heat of vaporization of water is 539 cal/kg.

=> q1 -> heat absobed in vaporisation = 0.0087 * 539 = 4.68 kcal

specific heat of water is 1 cal/g. K

=>q2 -> heat used in heating water 130g fom 19 to 100 = 130 * 1 * 81 /1000 = 10.53 kcal

specific heat of copper is .0923 cal/g.K

=> q3 = 130* 0.0923 * 91 /1000 = 0.012 kcal

Thus part(a) total energy to water as heat = q2 + q1 = 15.21 kcal

part(b) heat transfered to the bowl = q3 = 0.012 kcal or 12 cal

part(c) conserving heat -> q lost = m C dT =q1 + q2 + q3 = 15.222 kcal

=> (T - 100 ) = 15222/(0.0923 * 410 ) = 402.24 -> T = 502.24 K = 229.23 ~ 230 oC

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