Question

A piece of metal weighing $100\, g$ is heated to $80^{\circ} C$ and dropped into $1\, kg$ of cold water in an insulated container at $15^{\circ} C$. If the final temperature of the water in the container is $15.69^{\circ} C$, the specific heat of the metal in $J / g .{ }^{\circ} C$ is

• A. $0.38$
• B. $0.24$
• C. $0.45$
• D. $0.13$

Answer 1

Answer: (C)

Given,
weight of metal $=100\, g$
$T_{2}=15.69^{\circ} C,\, T_{1}=80^{\circ} C$
Specific heat of water $=4.184\, J / g ^{\circ} C$
Heat gained by $100\, g$ of metal $=$ Heat lost by $1000\, g$ of water.
We know, $Q=m c \Delta T$
$\therefore 100 \times x \times(80-15.69)$
$=1000 \times 4.184(15.69-15)$
$=100 x(64.31)=2886.96$
$x =\frac{2886.96}{6431}=0.448 \approx 0.45\, J / g .{ }^{\circ} C$