Question

The specific heat of substance varies with temperature according to equation c=(2t²+t)x 10^-3cal /gm°c   The amount of heat required to raise the temperature of 100 gm of substance from 20°c to 40°c  will be​

Answer 1

Given:

Mass of the substance = 100 gm

Initial temperature(T₁) = 20°C

Final temperature(T₂) = 40°C

c = (2t²+t)×10⁻³ Cal/gm°C

To find:

The amount of heat required to raise the temperature of the 100 gm of the substance

Calculation:

We know the relation

          $dQ=mcdt$

Substitute the given values in the above relation

          $dQ=[100\times(2t^{2}+t)\times10^{-3}]dt$

          $Q=\int\limits^{40}_{20} {100\times(2t^{2}+t)\times10^{-3}} \, dt$

          $Q=\frac{1}{10} \int\limits^{40}_{20} {(2t^{2}+t)} \, dt$

          $Q=\frac{1}{10} [\frac{2\times(40^3-20^3)}{3}}+\frac{(40^2-20^2)}{2} ]$

          $Q=3793.3\ Cal$

Final answer:

The amount of heat required to raise the temperature is 3793.3 Cal.