Question

A piece of copper of mass 2g is dropped in the glass of water. Initial temperature of copper and water is ${85}^{o}C$ and ${70}^{o}C$ respectively. Calculate the final temperature of the mixture given that mass of water is 12g and specific heat capacity of copper and water is $c_{Cu}=0.092\frac{cal}{g^{o}C}$ and $c_{water}=1.00\frac{cal}{g^{o}C}$ respectively.

• A. ${75.22}^{o}C$
• B. ${70.23}^{o}C$
• C. ${83.76}^{o}C$
• D. ${69.77}^{o}C$
• E. ${79.99}^{o}C$

Answer 1

The correct option is B ${70.23}^{o}C$

Let the attained final temperature of the mixture be $T$.

Thus the loss in heat energy of the copper piece=$m_{Cu}{c}_{Cu}\Delta T_{Cu}$

$=2\times 0.092\times (85-T)$

From conservation of energy, this loss must be equal to gain in heat energy of water=$m_{water}{c}_{water}\Delta T_{water}$

$=12\times 1.00\times (T-70)=2\times 0.092\times (85-T)$

$⟹T={70.23}^{\circ }C$