Question

A 2.0 -kg metal object with a temperature of \(90^{\circ} \mathrm{C}\) is submerged in \(1.0 \mathrm{~kg}\) of water at \(20^{\circ} \mathrm{C} .\) The water-metal system reaches equilibrium at \(32^{\circ} \mathrm{C} .\) What is the specific heat of the metal? a) \(0.840 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) b) \(0.129 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) c) \(0.512 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) d) \(0.433 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\)

Short answer

$$c_m = \frac{50.16}{116}$$ $$c_m \approx 0.43 \frac{\mathrm{kJ}}{\mathrm{kg·K}}$$ Based on the given information and calculations, the specific heat of the metal object is approximately 0.43 kJ/(kg·K).

Step-by-step solution

List the known and unknown values.

The known values are: - Mass of metal, \(m_m = 2.0 \,\text{kg}\). - Initial temperature of metal, \(T_{m,i} = 90^{\circ} \mathrm{C}\). - Mass of water, \(m_w = 1.0 \,\text{kg}\). - Initial temperature of water, \(T_{w,i} = 20^{\circ} \mathrm{C}\). - Final equilibrium temperature, \(T_f = 32^{\circ} \mathrm{C}\). - Specific heat of water, \(c_w = 4.18 \frac{\mathrm{kJ}}{\mathrm{kg·K}}\). The unknown value is the specific heat of the metal, \(c_m\).

Set up the heat exchange equation.

Write the heat exchange equation, which states that the heat lost by the metal is equal to the heat gained by the water: $$Q_m = Q_w.$$ Now, use the formula \(Q = mc\Delta T\) for both the metal and the water: $$m_mc_m(T_{m,i} - T_f) = m_wc_w(T_f - T_{w,i}).$$

Plug in the known values and solve for c_m.

Enter the known values into the equation from Step 2 and solve for \(c_m\): $$2c_m(90 - 32) = 1 \cdot 4.18(32 - 20).$$ Now, solve for \(c_m\): $$c_m = \frac{1 \times 4.18(12)}{2 \times 58}.$$