In an experiment, a sphere of aluminium of mass 0.20 kg is heated up to 150°C. Immediately, it is put into water of volume 150 cc at 27°C kept in a calorimeter of water equivalent to 0.025 kg. The final temperature of the system is 40°C. The specific heat of the aluminium is (take 4.2 Joule = 1 calorie):

We have to use the principle of calorimetry, namely that in an isolated system

$$\text{Heat lost by the hot body}= \text{Heat gained by the cold bodies}.$$

The heat absorbed or released by any body is given by the well-known formula

$$Q = m\,c\,\Delta T,$$

where $$m$$ is the mass, $$c$$ is the specific heat capacity and $$\Delta T$$ is the change in temperature.

Data for the aluminium sphere

Mass $$m_1 = 0.20\;\text{kg}$$, initial temperature $$T_1 = 150^{\circ}\text{C}$$, final temperature $$T_f = 40^{\circ}\text{C}$$.

So the temperature fall is

$$\Delta T_1 = T_1 - T_f = 150^{\circ}\text{C} - 40^{\circ}\text{C} = 110^{\circ}\text{C}.$$

Data for the water in the calorimeter

Volume given is $$150\;\text{cc}.$$ Because the density of water is $$1\;\text{g cm}^{-3},$$ the mass is

$$m_2 = 150\;\text{g} = 0.150\;\text{kg}.$$

Initial temperature $$T_2 = 27^{\circ}\text{C},$$ final temperature $$T_f = 40^{\circ}\text{C}$$; hence

$$\Delta T_2 = T_f - T_2 = 40^{\circ}\text{C} - 27^{\circ}\text{C} = 13^{\circ}\text{C}.$$

Data for the calorimeter itself

The water equivalent is $$0.025\;\text{kg}$$. This means its heat capacity is the same as that of $$0.025\;\text{kg}$$ of water. Therefore we can treat it as additional water of mass

$$m_3 = 0.025\;\text{kg},$$

with the same temperature rise

$$\Delta T_3 = \Delta T_2 = 13^{\circ}\text{C}.$$

Specific heat of water

For water we use $$c_w = 4200\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.$$ (The conversion given, $$4.2\;\text{J} = 1\;\text{cal},$$ is equivalent to this value.)

Heat gained by water

$$Q_2 = m_2\,c_w\,\Delta T_2 = 0.150\;\text{kg}\times 4200\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}\times 13^{\circ}\text{C}.$$

First multiply the mass and specific heat:

$$0.150 \times 4200 = 630\;\text{J}\,{}^{\circ}\text{C}^{-1}.$$

Then multiply by the temperature rise:

$$630 \times 13 = 8190\;\text{J}.$$

Heat gained by the calorimeter

$$Q_3 = m_3\,c_w\,\Delta T_3 = 0.025\;\text{kg}\times 4200\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}\times 13^{\circ}\text{C}.$$

Calculating similarly:

$$0.025 \times 4200 = 105\;\text{J}\,{}^{\circ}\text{C}^{-1},$$

and

$$105 \times 13 = 1365\;\text{J}.$$

Total heat gained by water plus calorimeter

$$Q_{\text{gained}} = Q_2 + Q_3 = 8190\;\text{J} + 1365\;\text{J} = 9555\;\text{J}.$$

Heat lost by the aluminium sphere

Let $$c$$ be the specific heat capacity of aluminium (in $$\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$$). Then

$$Q_{\text{lost}} = m_1\,c\,\Delta T_1 = 0.20\;\text{kg}\times c \times 110^{\circ}\text{C}.$$

Multiplying the numerical factors first:

$$0.20 \times 110 = 22,$$

so

$$Q_{\text{lost}} = 22\,c.$$

Applying the heat balance

Because no heat escapes the system,

$$Q_{\text{lost}} = Q_{\text{gained}}.$$

Therefore

$$22\,c = 9555.$$

Solving for $$c$$

$$c = \frac{9555}{22}\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.$$

Carrying out the division:

$$c = 434.318\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}\;(\text{approximately}).$$

Rounding to three significant figures gives

$$c \approx 434\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.$$

Hence, the correct answer is Option A.