Water of mass m at $30^{\circ} \mathrm{C}$ is mixed with 5 g of ice at $-20^{\circ} \mathrm{C}$. If the…

Water of mass m at $30^{\circ} \mathrm{C}$ is mixed with 5 g of ice at $-20^{\circ} \mathrm{C}$. If the resultant temperature of the mixture is $6^{\circ} \mathrm{C}$, then the value of $m$ is (specific heat capacity of ice $\mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ )
  1. 48 g
  2. 20 g
  3. 24 g
  4. 40 g

Solution

By principle of calorimetry, $\begin{aligned} & \text { Heat lost = Heat gained } \\ & \Rightarrow \mathrm{m}_{\mathrm{w}} \mathrm{C}_{\mathrm{pw}}\left(\mathrm{T}_{\mathrm{w}}-\mathrm{T}\right)=\mathrm{m}_{\mathrm{i}} \mathrm{c}_{\mathrm{pi}}\left(\mathrm{o}-\mathrm{T}_{\mathrm{i}}\right)+\mathrm{m}_{\mathrm{i}} \mathrm{L}_{\mathrm{i}}+\mathrm{m}_{\mathrm{i}} \\ & \mathrm{c}_{\mathrm{pw}}(\mathrm{T}-\mathrm{O}) \\ & \Rightarrow \mathrm{m} \times 1 \times(30-6)=5 \times 0.5(0+20)+5 \times 80+5 \times 1 \\ & \times(6-0) \\ & \therefore \mathrm{m}=\frac{50+400+30}{24}=20 \mathrm{~g}\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

Practice more Thermal Properties of Matter questions on Aicharya