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Question

A 50 g sample of metal was heated to 1000C and then dropped into a beaker containing 50g of water at 250C. If the specific heat capacity of the metal is 0.25 cal/g, what is the final temperature of the water in 0C?

A
27
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B
40
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C
60
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D
86
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Solution

The correct option is B 40
[mC(TfTi)]metal+[mC(TfTi)]water=0
Which can then become:

[mC(TiTf)]metal=[mC(TfTi)]water
50×0.25(100Tf)=50×4.184(Tf25)
12.5(100Tf)=209.2(Tf25)
125012.5Tf=209.2Tf5230
1250+5230=209.2Tf+12.5Tf
6480=221.7Tf
Tf=6480221.7=29oC


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