A 800 g block of metal at 100 °C is dropped into 1.5 kg of water at 15 °C. The final temperature of the mixture is 20 °C. What is the specific heat capacity of the metal?

Given:

• Mass of metal, $m_m = 800 \, \text{g} = 0.8 \, \text{kg}$
• Initial temperature of metal, $T_{m,i} = 100 \, \text{°C}$
• Mass of water, $m_w = 1.5 \, \text{kg}$
• Initial temperature of water, $T_{w,i} = 15 \, \text{°C}$
• Final temperature of the mixture, $T_f = 20 \, \text{°C}$
• Specific heat capacity of water, $c_w = 4200 \, \text{J kg}^{-1} \, \text{°C}^{-1}$ (standard value)

Unknown:

• Specific heat capacity of metal, $c_m$

Change in temperature of metal:

$$\Delta T_m = T_{m,i} - T_f = 100 \, \text{°C} - 20 \, \text{°C} = 80 \, \text{°C}$$

Change in temperature of water:

$$\Delta T_w = T_f - T_{w,i} = 20 \, \text{°C} - 15 \, \text{°C} = 5 \, \text{°C}$$

According to the principle of calorimetry, the heat lost by the hot body (metal) is equal to the heat gained by the cold body (water), assuming no heat loss to the surroundings.

$$\text{Heat lost by metal} = \text{Heat gained by water}$$ $$Q_m = Q_w$$

The formula for heat transfer is $Q = mc\Delta T$.

$$m_m c_m \Delta T_m = m_w c_w \Delta T_w$$

$$(0.8 \, \text{kg}) \times c_m \times (80 \, \text{°C}) = (1.5 \, \text{kg}) \times (4200 \, \text{J kg}^{-1} \, \text{°C}^{-1}) \times (5 \, \text{°C})$$ $$64 \, c_m = 31500$$ $$c_m = \frac{31500}{64}$$ $$c_m = 492.1875 \, \text{J kg}^{-1} \, \text{°C}^{-1}$$

The specific heat capacity of the metal is $\text{492.19 J kg}^{-1} \text{°C}^{-1}$.