Question

Solid whose specific heat is to be determined is heated uniformly up to a fixed temperature in a hypsometer. It is then dropped in a calorimeter containing known quantity of cold water and the temperature of mixture is recorded. The hot solid will lose heat whereas calorimeter and cold water will gain heat till the temperature becomes same throughout.

If $t$ and $\theta$ be the temperature of cold water and temperature of mixture finally respectively, then the total heat gained by calorimeter, stirrer and cold water is given by ($W=$ water equivalent of calorimeter and $m=$ mass of cold water)

• A. $(W+m)(t-\theta )$
• B. $(W+m)(\theta -t)$
• C. $(W+m)(\theta +t)$
• D. $(\theta +t)(m-W)$

Answer 1

Answer: (B)

$(W+m)(\theta -t)$

Heat gain by calorimeter, $Q_1=W(\theta -t)\times 1$

Heat gain by cold water, $Q_2=m(\theta -t)\times 1$

Total heat is $Q_1+Q_2=(W+m)(\theta -t)$