Question

A piece of iron weighing 16g at temperature of ${112.5}^{0}C$ is dropped over a block of ice 2.5 g of ice melts. If the latent heat of ice is 80 cal/g, then the specific heat of iron is :

• A. 1 Cal $g^{-1}{C}^{-1}$
• B. 1/9 Cal $g^{-1}{C}^{-1}$
• C. 0.01 Cal $g^{-1}{C}^{-1}$
• D. 1/16 Cal $g^{-1}{C}^{-1}$

Answer 1

Answer: (B)

1/9 Cal $g^{-1}{C}^{-1}$

Let the specific heat be $cCal{g}^{-1}{C}^{-1}$.

So heat released by iron is $c\times 16\times 112.5Cal$.

Heat absorbed by melted ice is $2.5\times 80=200Cal$

Balancing the heat released and heat absorbed:

$c=200/(16\times 112.5)=1/9Cal{g}^{-1}{C}^{-1}$