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Question

The piece of zinc at a temperature of 20oC weighing 65.38 g is dropped into 180 g of boiling water (T=100oC). The specific heat of zinc is 0.400 Jg−1oC−1 and that of water is 4.20 Jg−1oC−1. What is the final common temperature reached by both the zinc and water :-

A
97.3oC
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B
33.4oC
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C
80.1oC
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D
60oC
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Solution

The correct option is A 97.3oC
mZn.SZn.(TfTi)+mH2O.SH2O.(TfTi)=0

(65.38gm)(0.4J/goC)(Tf20oC)+180gm(4.20J/goC)×(Tf100oC)=0

[(65.38)(0.4)+180(420)]Tf=(65.38)(0.4)(20)+(180)(4.20)(100)

Tf=[(65.38)(0.4)(20)+(180)(4.20)(100)(65.38)(0.4)+(180)(4.20)]=97.3oC

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