A 110 g copper bowl contains 200 g of water, both at 18.0 degree. A very hot 440

Question

A 110 g copper bowl contains 200 g of water, both at 18.0 degree. A very hot 440 g copper cylinder is dropped into the water, causing the water to boil, with 6.36 g being converted to steam, The final temperature of the system is 100 degree C, Neglect energy transfers with the environment (a) How much energy is transformed 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 middot K, and of copper is 0.0923 cal/g middot K. The heat of water is 539

Explanation / Answer

a)

Heat gained by bowl = ms(T2-T1)=110 *0.0923* [100 - 18] = 832.546 cal.
Heat gained by 200 gm of water = 200 *1*[100-18] =16400 cal.
Heat used to convert 6.36 g of water into steam = mL
= 539*6.36 = 3428.04cal.
energy transferred to the water is=16400 cal.+3428.04=19828.40 cal

so the answer is 19828cal(in five significant figure)

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b)

Heat gained by bowl = ms(T2-T1)=110 *0.0923* [100 - 18] = 832.546 cal.

so the answer is 832.55cal

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c)

Total calories gained by water and bowl=16400+3428.40+832.546=20660.586cal

Heat lost by440g of copper is 440* 0.0923*[T - 100]
= 40.612[T - 100] cal.=20660.586cal

T-100=20660.586cal/46.612=508.73

T=608.730C

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