Two bodies of same material having masses m1 and m2 are supplied equal amount of energy

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Text Solution

Solution : For body A, `Q=mxxs_(A)xxDeltatheta_(A)` <br> and for body B, `Q=mxxs_(B)xxDeltatheta_(B)` <br> Dividing (1) by (2), `(S_(A))/(S_(B))=(Deltatheta_(A))/(Deltatheta_(B))` <br> This shows that specific heat is inversely proportional to rise in temperature, Thus, specific heat of A is less than that of B.

Text Solution

Solution : Given heat, H = mst i.e., temperature rise, `t=(H)/(ms)` <br> Since for the bodies, heat supplied H and mass m are equal, `t prop (1)/(s)`. <br> Hence, the body with comparatively less specific heat will have a higher rise in temperature.

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