91Ó°ÊÓ

1. The density of the liquid is$\mathrm{ÒÏ}=0.85\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ .
2. The volume of the liquid flowing through calorimeter per second is$V=8.0\frac{{\text{cm}}^{\text{3}}}{\text{s}}$.
3. The rate of adding or takeaway of energy from the calorimeter is$P=250\text{W}$.
4. The temperature difference between the inflow and outflow is $\mathrm{Δ}T=15\text{°°ä}$.

The amount of heat required to increase the temperature of 1 gram of a substance by 1 degree Celsius is known as its specific heat. We can use the concept of specific heat of the water. We can use the expression of density and find the mass of the liquid. By using the concept of power, we can find the amount of heat that this mass takes away from the calorimeter.

Formulae:

Density of the body,$\mathrm{ÒÏ}=\frac{m}{V}$…(¾±)

Rate of radiation by the body via thermal radiation,$P=\frac{Q}{t}$…(¾±¾±)

Heat transferred by the body via thermal energy,$Q=cm(T_f−T_i)$ .…(¾±¾±¾±)

The system is in a steady state; hence, the amount of heat added to the liquid, in one second, from the heating coil is equal to the amount of heat flowing away from the liquid in one second through the calorimeter.

The volume of the liquid flowing through the calorimeter in one second is$V=8.0{\text{cm}}^{\text{3}}$.

From equation (i), the value of the mass of the body is given by

$\begin{array}{c}m=0.85\frac{\text{g}}{{\text{cm}}^{\text{3}}}×8.0{\text{cm}}^{\text{3}}\\ =6.8\text{g}\end{array}$

In one second, the amount of heat that this mass takes away from the calorimeter using equation (ii) is given by

$\begin{array}{c}Q=250\text{W}×1\text{s}\\ =250\text{J}\end{array}$

The expression for the specific heat of the liquid using equation (iii) is given by

$\begin{array}{c}c=\frac{Q}{m\mathrm{Δ}T}\\ =\frac{250\text{J}}{6.8\text{g}×15\text{°°ä}}\\ =2.45\frac{\text{J}}{\text{g°°ä}}\\ ≈2.5×{10}^3\frac{\text{J}}{\text{°ì²µÂ°°ä}}\end{array}$

Hence, the value of the specific heat of the liquid is $2.5×{10}^3\frac{\text{J}}{\text{°ì²µÂ°°ä}}$.