Question

$10\, g$ of ice at $0^{\circ}C$ is mixed with $100\,g$ of water at $50^{\circ}C$ in a calorimeter. The final temperature of the mixture is [Specific heat of water $= 1\, cal\, g^{-1}\,{}^{\circ}C^{-1}$, latent heat of fusion of ice $= 80 \,cal\, g^{-1}$]

• A. $31.2^{\circ}C$
• B. $32.8^{\circ}C$
• C. $36.7^{\circ}C$
• D. $38.2^{\circ}C$

Answer 1

Answer: (D)

Here,
Mass of water, $m_w = 100 \,g$
Mass of ice, $m_i= 10\, g$
Specific heat of water, $s_{w}=1\,cal\,g^{-1}\,{}^{\circ}C^{-1}$
Latent heat of fusion of ice, $L_{fi} = 80\, cal\, g^{-1}$
Let $T$ he the final temperature of the mixture.
Amount of heat lost by water
$=m_{w}s_{w}\left(\Delta T\right)_{w}=100 \times 1 \times\left(50-T\right)$
Amount of heat lost by water
$=m_{i}L_{fi}+m_{i}s_{w}\left(\Delta T\right)_{i}$
$=10 \times 80 +10 \times1\times\left(T-0\right)$
According to principle of calorimetry
Heat lost $=$ Heat gained
$100 \times 1 \times (50 - T) $
$= 1 0 \times 8 0 + 10 \times 1 \times (T -0)$
$500 - 10T = 80 + T$
$11T = 420$ or $7 =38.2\,{}^{\circ}C$