Question: A0.300 kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate of 2.81 W . Figure 18-52 gives the temperature Tof the sample versus time t.The temperature scale is set by ${\mathit{T}}_{\mathbf{s}}\mathbf{=}\mathbf{30}\mathbf{°}\mathbf{\text{C}}$and the time scale is set by ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{20}\mathit{m}\mathit{i}\mathit{n}$.What is the specific heat of the sample?

1. Mass of the sample is$m=0.300\text{kg}.$
2. Rate of removal of heat is$P=2.81\text{W}$
3. Temperature vs. time graph is given.
4. Time scale is set at,$t_s=20\min \text{or}1200\text{s}$
5. Temperature scale is set at $T_s={30}^0\text{C}$

In the cooling apparatus, when a substance is placed it is brought to a lower temperature than its present temperature at a certain rate of radiation. Thus, the heat released due to the change in temperature at a certain rate within a given time is calculated, that is further used for the calculation of the specific heat of the substance.

Formula:

The heat transferred by the body due to rate of conduction, $Q=P\times t$ …(i)

Where,P is the amount of radiated power, is the time of radiation process.

The heat released by the body, $Q=mc\Delta T$ …(ii)

Where,m is the mass of the substance, c is the specific heat of the substance, $\Delta T$is the temperature difference.

Substituting the given values in equation (i), we can get the heat absorbed y the body in cooling down is given as:

$$\begin{gathered} Q=2.81\text{W}\times 1200\text{s} \\ =3372\text{W - s} \\ =3372\text{J} \end{gathered}$$

From equation (ii) and given values, the specific heat of the sample is given as:

$$\begin{gathered} c=\frac{Q}{m\Delta T} \\ =\frac{3372\text{J}}{0.300\text{kg}\left(25\text{K}\right)}\text{(}\because \Delta T=25\text{K},\text{from graph)} \\ =449.6\text{J/kgK} \\ \approx 4.5\times {10}^2\text{J/kgK} \end{gathered}$$