91Ó°ÊÓ

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, $Q=P×t$ …(¾±)

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

The heat released by the body, $Q=mc\mathrm{Δ}T$ …(¾±¾±)

Where,m is the mass of the substance, c is the specific heat of the substance, $\mathrm{Δ}T$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:

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

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

$$\begin{gathered} c=\frac{Q}{m\mathrm{Δ}T} \\ =\frac{3372\text{J}}{0.300\text{kg}\left(25\text{K}\right)}\text{(}âˆ\mu \mathrm{Δ}T=25\text{K},\text{from graph)} \\ =449.6\text{J/kgK} \\ ≈4.5×{10}^2\text{J/kgK} \end{gathered}$$