WBJEE · Physics · Thermal Properties of Matter
When \(100 \mathrm{~g}\) of boiling water at \(100^{\circ} \mathrm{C}\) is added into a calorimeter containing \(300 \mathrm{~g}\) of cold water at \(10^{\circ} \mathrm{C}\), temperature of the mixture becomes \(20^{\circ} \mathrm{C}\). Then a metallic block of mass \(1 \mathrm{~kg}\) at \(10^{\circ} \mathrm{C}\) is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes \(19^{\circ} \mathrm{C}\). What is the specific heat of the metal in C.G.S. unit?
- A \(0.01\)
- B \(0.3\)
- C \(0.09\)
- D \(0.1\)
Answer & Solution
Correct Answer
(D) \(0.1\)
Step-by-step Solution
Detailed explanation
Hint: Let, heat capacity of calorimeter = ms \(\begin{aligned} & 100 \times 1 \times 80=300 \times 1 \times 10+\mathrm{ms} \times 10 \\ \therefore \mathrm{ms}=500 \mathrm{Cal} /{ }^{\circ} \mathrm{C} \end{aligned}\) Again,…
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