AP EAMCET · PHYSICS · Thermal Properties of Matter
The time taken for a calorimeter containing \(75 \mathrm{~g}\) of water at \(62^{\circ} \mathrm{C}\) to \(\mathrm{cool}\) to \(58^{\circ} \mathrm{C}\) is 9 minutes. When the calorimeter contains \(105 \mathrm{~g}\) of water, it takes 12 minutes to cool from \(62^{\circ} \mathrm{C}\) to \(58^{\circ} \mathrm{C}\). The water equivalent of the calorimeter is
- A 10 g
- B 15 g
- C 20 g
- D 30 g
Answer & Solution
Correct Answer
(B) 15 g
Step-by-step Solution
Detailed explanation
Heat lost \(=(w+m) C_V d T\) where, \(w=\) water equivalent of calorimeter, \(m=\) mass of water, \(C_V=\) specific heat of water and \(\quad d T=\) temperature difference \(=T_2-T_1\). Time taken to cool down is proportional to amount of heat lost. Case I 9 minutes…
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