Q. A copper block of mass 4 kg is heated in a furnance to a temperature 425$℃$ and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper=500 J ${kg}^{- 1}$ $℃^{- 1}$ and heat of fusion of ice=336 k J ${kg}^{- 1}$)
A) 0.5 kg
B) 1 kg
C) 1.5 kg
D) 2.5 kg
Explanation
(d) Fall in temperature of copper block when it is placed on the ice block = $\mathrm{\Delta}T = 425 - 0 = 425℃.$ Heat lost by copper block when it is placed on the ice block.
$Q_{1} = m_{1}s\mathrm{\Delta}T$
= $4 \times 500 \times 425 = 850kJ$
Heat gained by ice in melting into $m_{2}$ kg of water.
$Q_{2} = m_{2}L$
$\ = m_{2} \times 336$
$= 336m_{2}\ kJ$
According to Calorimetry principle,
Heat lost=Heat gained
$ie,$ 850=$336m_{2}$
$\therefore$ $m_{2} = \frac{850}{336} = 2.5kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.