WBJEE · Physics · Thermal Properties of Matter
The water equivalent of a calorimeter is \(10 \mathrm{g}\) and it contains \(50 \mathrm{g}\) of water at \(15^{\circ} \mathrm{C}\). Some amount of ice, initially at \(-10^{\circ} \mathrm{C}\) 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 \(=0.5 \mathrm{cal} \mathrm{gm}^{-1}{ }^{\circ} \mathrm{C}^{-1},\) specific heat of water \(=1.0\)
cal \(\mathrm{gm}^{-1 \circ} \mathrm{C}^{-1}\) and latent heat of melting of ice \(=80\) cal \(\mathrm{gm}^{-1}\) )?
- A \(10 \mathrm{g}\)
- B \(18 \mathrm{g}\)
- C \(20 \mathrm{g}\)
- D \(30 \mathrm{g}\)
Answer & Solution
Correct Answer
(C) \(20 \mathrm{g}\)
Step-by-step Solution
Detailed explanation
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…
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