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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

  1. A 10 g
  2. B 15 g
  3. C 20 g
  4. D 30 g
Verified Solution

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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