A copper beaker of mass 0.20 kg contains 0.10 kg of water at 20.0 celsius. A blo
ID: 2123001 • Letter: A
Question
A copper beaker of mass 0.20 kg contains 0.10 kg of water at 20.0 celsius. A block of silver of mass 100 g at 90.0 celsius is added to the beaker of water. The entire system is thermally insulated from the surroundings.(specific heats-copper: 390J/Kg K, silver: 234 J/ Kg K, water: 4190 J/Kg K)
(a) What is the final temperature of the system?
(b) What is the heat lost by the silver?
(c) What is the heat gained by the water? What is the heat gain of the copper?
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Explanation / Answer
Mc=.20kg Mw=.10Kg T1=20+273=293K
Ms=0.10Kg T2=90+273=363K
Specific Heat of copper=390 J/Kg K
Specific Heat of silver=234 J/Kg K
Specific Heat of water=4190 J/Kg K
let final temperature is T then
Energy given by silver =Energy taken by(copper +water)
0.10*234*(363-T)=(T-293)(0.20*390+0.1*4190)
23.4(363-T)=497(T-293)
520.4 T= 154115.2
T=296.14 K=296.14-273=23.14 degree centigrade
b)Heat lost by silver=Ms *specific heat of silver*temperature difference
=0.10*234*(363-296.14)
=23.4*66.86
=1564.524 J
c)Heat gained by copper=Mc *specific heat of copper*temerature difference
=0.20*390*(296.14-293)
=7800*3.14
=244.92 J
Heat gained by water=Mw *specific heat of water*temerature difference
=0.10*4190*3.14
=1315.66 J
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