A metal ball of mass 100 g at $20^{\circ} \mathrm{C}$ is dropped in 200 ml of water at $80^{\circ}…

A metal ball of mass 100 g at $20^{\circ} \mathrm{C}$ is dropped in 200 ml of water at $80^{\circ} \mathrm{C}$. If the resultant temperature is $70^{\circ} \mathrm{C}$, then the ratio of specific heat of the metal to that of water is
  1. $\frac{5}{2}$
  2. $\frac{1}{2}$
  3. $\frac{2}{5}$
  4. $\frac{2}{1}$

Solution

By principle of calorimetry
Heat gained = Heat lost $\begin{aligned} & \Rightarrow \mathrm{m}_1 \mathrm{~s}_1\left(\mathrm{~T}-\mathrm{T}_1\right)=\mathrm{m}_2 \mathrm{~s}_2\left(\mathrm{~T}_2-\mathrm{T}\right) \\ & \Rightarrow 100 \times \mathrm{s}_1 \times(70-20)=200 \times \mathrm{S}_2(80-70) \\ & \therefore \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{200 \times 10}{100 \times 50}=\frac{2}{5} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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