Question

An ice cube of mass $0.1 \,kg$ at $0^{\circ}C$ is placed in an isolated container which is at $227^{\circ}C$. The specific heat s of the container varies with temperature $T$ according to the empirical relation $s = A + BT$, where $A = 100\, cal\, kg^{-1}\, K^{-1}$ and $B = 2 \times 10^{-2}\, cal\, kg^{-1}$. If the final temperature of the container is $27^{\circ}C$, the mass of the container is (Latent heat of fusion of water $= 8 \times 10^4\, cal\, kg^{-1}$, specific heat of water $= 10^3\,cal\, kg^{-1}\, K^{-1}$)

• A. $0.495\, kg$
• B. $0.595\, kg$
• C. $0.695\, kg$
• D. $0.795 \,kg$

Answer 1

Answer: (A)

$0.495\, kg$

Answer 2

Heat lost by container $=-\int \limits^{300}_{500} m_{C}\left(A+BT\right)dT$ $=-m_{C}\left[AT+\frac{BT^{2}}{2}\right]^{300}_{500}=21600\,m_{C}$ Heat gained by ice $=m_{ice}L+m_{ice}s_{water}\,\Delta T$ $=0.1 \times8 \times 10^{4}+0.1 \times10^{3}\times 27$ $=10700\,cal$ According to principle of calorimetry Heat lost by container $=$ Heat gained by ice $21600m_C = 10700$ or $m_C=0.495\,kg$