Chapter 18: Q34P (page 542)

Samples A and B are at different initial temperatures when they are placed in a thermally insulated container and allowed to come to thermal equilibrium. Figure a gives their temperatures T versus time t. Sample A has a mass of $5 .0 kg$ ; sample B has a mass of $1 .5 kg$ . Figure b is a general plot for the material of sample B. It shows the temperature change $Δ T$ that the material undergoes when energy is transferred to it as heat Q. The change $Δ T$ is plotted versus the energy Q per unit mass of the material, and the scale of the vertical axis is set by $T_{s} = 4 {.0}^{0} C$ .What is the specific heat of sample A?

Short Answer

Expert verified

The specific heat of sample A is $4 \times 10^{2} \frac{J}{{kg.}^{0} C}$

Step by step solution

The given data

From fig.18-34(b) we have

i) Reciprocal of the slope is $c_{B} = \frac{16}{4} = 4 \frac{kJ}{{kg.}^{0} C}$

ii) Specific heat of sample B is $c_{B} = 4000 \frac{J}{{kg.}^{0} C}$

iii) Mass of sample A is $(m_{A}) = 5.0 kg$

iv) Mass of the sample B is $(M_{B}) = 1.5 kg$

v) Temperature of sample A is $(T_{A}) = 100^{0} C$

vi) Temperature of the sample B is $(T_{B}) = 20^{0} C$

vii) Final temperature of the whole system is $(T_{f}) = 40^{0} C$

Understanding the concept of specific heat

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the formula for specific heat capacity and the concept of equilibrium to find the specific heat capacity. The amount of heat lost by sample A should be equal to the amount of heat gained by sample B. The exchange of heat would continue till both the samples reach equilibrium or achieve the same temperature.

Formula:

The heat energy required by a body, $Q = m c Δ T$ …(i)

Where, $m$ = mass

$c$ = specific heat capacity

$Δ T$ = change in temperature

$Q$ = required heat energy

Calculation of the specific heat of sample A

If sample A and B are placed in a thermally insulated conductor and allowed to come to thermal equilibrium, then we can write,

$Q_{A} + Q_{B} = 0$ …(ii)

Substituting equation (i) value for the heat of samples A and B in equation (ii), we can get the specific heat of sample A is given as:

$\begin{matrix} m_{A} c_{A} (T_{f} - T_{A}) + m_{B} c_{B} (T_{f} - T_{B}) = 0 \\ m_{A} c_{A} (T_{f} - T_{A}) = - m_{B} c_{B} (T_{f} - T_{B}) \\ c_{A} = \frac{- m_{B} c_{B} (T_{f} - T_{B})}{m_{A} (T_{f} - T_{A})} \\ = \frac{- 1.5 kg \times 4000 \frac{J}{{kg.}^{0} C} \times {(40 - 20)}^{0} C}{5 kg \cdot {(40 - 100)}^{0} C} \\ = \frac{- 120000 J}{- 300 {kg}^{0} C} \\ = 400 \frac{J}{{kg.}^{0} C} \\ = 4 \times 10^{2} \frac{J}{{kg.}^{0} C} \end{matrix}$