Question

A metal block is made from a mixture of 2.4 kg of aluminium, 1.6 kg of brass, and 0.8 kg of copper. The metal block is initially at 20°C. If the heat supplied to the metal block is 44.4 calories, find the final temperature of the block if specific heats of aluminium, brass, and copper are 0.216, 0.0917, and 0.0931 cal.kg\(^{-1}\)°C\(^{-1}\) respectively.

• A. \( 100^\circ C \)
• B. \( 60^\circ C \)
• C. \( 40^\circ C \)
• D. \( 80^\circ C \)

Hint:

The total heat capacity of a mixed system is found using the weighted sum of individual heat capacities.

Answer 1

The Correct Option is D

Step 1: Heat Energy Equation The heat absorbed by the metal block is given by: \[ Q = mc\Delta T \] where: - \( Q \) = heat energy supplied = 44.4 cal
- \( m \) = mass of the metal
- \( c \) = specific heat of the metal
- \( \Delta T \) = change in temperature Step 2: Calculating Effective Specific Heat Capacity The total heat capacity of the metal block is: \[ C_{\text{eff}} = (m_1 c_1 + m_2 c_2 + m_3 c_3) \] Substituting values: \[ C_{\text{eff}} = (2.4 \times 0.216) + (1.6 \times 0.0917) + (0.8 \times 0.0931) \] \[ C_{\text{eff}} = 0.5184 + 0.14672 + 0.07448 = 0.7396 \text{ cal/°C} \] Step 3: Finding Final Temperature Using the equation: \[ Q = C_{\text{eff}} \times (T_f - T_i) \] \[ 44.4 = 0.7396 \times (T_f - 20) \] Solving for \( T_f \): \[ T_f - 20 = \frac{44.4}{0.7396} = 60 \] \[ T_f = 80^\circ C \] Thus, the correct answer is option (4).

Answer 2

Given:
- Mass of aluminium, \( m_{Al} = 2.4 \, \text{kg} \)
- Mass of brass, \( m_{Brass} = 1.6 \, \text{kg} \)
- Mass of copper, \( m_{Cu} = 0.8 \, \text{kg} \)
- Initial temperature, \( T_i = 20^\circ C \)
- Heat supplied, \( Q = 44.4 \, \text{cal} \)
- Specific heat capacities:
- Aluminium, \( c_{Al} = 0.216 \, \text{cal} \cdot \text{kg}^{-1} \cdot {}^\circ C^{-1} \)
- Brass, \( c_{Brass} = 0.0917 \, \text{cal} \cdot \text{kg}^{-1} \cdot {}^\circ C^{-1} \)
- Copper, \( c_{Cu} = 0.0931 \, \text{cal} \cdot \text{kg}^{-1} \cdot {}^\circ C^{-1} \)

Find: Final temperature \( T_f \).

Step 1: Total heat supplied is absorbed by all three metals causing temperature rise:
\[ Q = m_{Al} c_{Al} (T_f - T_i) + m_{Brass} c_{Brass} (T_f - T_i) + m_{Cu} c_{Cu} (T_f - T_i) \]

Step 2: Factor out \( (T_f - T_i) \):
\[ Q = (T_f - T_i) \left( m_{Al} c_{Al} + m_{Brass} c_{Brass} + m_{Cu} c_{Cu} \right) \]

Step 3: Calculate the sum inside the parenthesis:
\[ m_{Al} c_{Al} = 2.4 \times 0.216 = 0.5184 \] \[ m_{Brass} c_{Brass} = 1.6 \times 0.0917 = 0.14672 \] \[ m_{Cu} c_{Cu} = 0.8 \times 0.0931 = 0.07448 \] Sum: \[ 0.5184 + 0.14672 + 0.07448 = 0.7396 \]

Step 4: Substitute values in heat equation:
\[ 44.4 = (T_f - 20) \times 0.7396 \] \[ T_f - 20 = \frac{44.4}{0.7396} \approx 60 \] \[ T_f = 20 + 60 = 80^\circ C \]

Therefore, the final temperature of the metal block is:
\[ \boxed{80^\circ C} \]