Question

A piece of zinc at a temperature of ${20}^{o}C$ weighing 65.38 g is dropped into 180 g of boiling water $(T={100}^{o}C).$ The specific heat of zinc is $0.4J{g}^{-1\circ }{C}^{-1}$ and that of water is $4.2J{g}^{-1o}{C}^{-1}.$ What is the final common temperature reached by both the zinc and water?

• A. $97.3{}^{o}C$
• B. $33.4{}^{o}C$
• C. $80.1{}^{o}C$
• D. $60.0{}^{o}C$

Answer 1

The correct option is A $97.3{}^{o}C$

Let final common temperature is $T_f$
According to law of energy conservation :
Heat gained by Zinc piece = Heat lost by water
$m_{Zn}s△T=m_{H_{2}O}s△T$, where, s is specific heat capacity.
$65.38\times 0.4(T_f-293)=180\times 4.2(373-T_f)$
$=\frac{T_f-293}{373-T_f}=28.90$
$=29.90T_f=28.90\times 373+293$
$\Rightarrow T_f=370.32K$
Hence, final common temperature reached by both the zinc and water $\left(T_f\right)={97.3}^{o}C$