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  4. The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15°C. Some amount of ice, initially at -10°C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal gm-1°C-1, specific heat of water = 1.0 cal gm-1°C-1 and latent heat of melting of ice = 80 cal gm-1) ?

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Q. The water equivalent of a calorimeter is $10\, g$ and it contains $50 \,g$ of water at $15°C$. Some amount of ice, initially at $-10°C$ is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice $= 0.5 \,cal\, gm^{-1}°C^{-1},$ specific heat of water $= 1.0 \,cal\, gm^{-1}°C^{-1}$ and latent heat of melting of ice $= 80 \,cal \,gm^{-1}$) ?

WBJEEWBJEE 2018Thermal Properties of Matter

Solution:

Let the mass of ice $=m$
Applying calorimetry principle,
heat given $=$ heat taken
$\left(m_{1}+m_{2}\right) s_{1}\left(t_{1}-t\right)=\frac{m L}{2}+m s_{2}\left(t-t_{2}\right)$
Putting values, we get
$(10+50) \times 1 \times(15-0)=\frac{m}{2} \times 80+m \times 0.5[0-(-10)]$
$\Rightarrow 60 \times 15=40 m+\frac{m}{2} \times 10=45\, m$
$\therefore m=\frac{60 \times 15}{45}=20\, g$