A metal brick is made from a mixture of 2.4 kg of aluminum 1.6 kg of brass and 0.8 kg of copper. The amount of heat required to raise the temperature of this block from `20^(@)C` to `80^(@)C` is (specific heats of aluminum brass and copper are `0.216, 0.0917, 0.0923 cal kg^(-1)``^(@)C^(-1)` respectively)

Options

A. 96.2 cal
B. 44.4 cal
C. 86.2 cal
D. 62.8 cal

Explanation

To solve the problem of calculating the amount of heat required to raise the temperature of a metal brick made from a mixture of aluminum, brass, and copper, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of aluminum (m₁) = 2.4 kg
- Mass of brass (m₂) = 1.6 kg
- Mass of copper (m₃) = 0.8 kg
- Specific heat of aluminum (c₁) = 0.216 cal/kg°C
- Specific heat of brass (c₂) = 0.0917 cal/kg°C
- Specific heat of copper (c₃) = 0.0923 cal/kg°C
- Initial temperature (T₁) = 20°C
- Final temperature (T₂) = 80°C

2. **Calculate the Change in Temperature (ΔT):**
\[ \Delta T = T_2 - T_1 = 80°C - 20°C = 60°C \]

3. **Use the Formula for Heat (Q):**
The heat required to raise the temperature of each material can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Therefore, the total heat (Q_net) required for the entire brick is:
\[ Q_{net} = (m_1 \cdot c_1 + m_2 \cdot c_2 + m_3 \cdot c_3) \cdot \Delta T \]

4. **Substitute the Values:**
\[ Q_{net} = \left( (2.4 \, \text{kg} \cdot 0.216 \, \text{cal/kg°C}) + (1.6 \, \text{kg} \cdot 0.0917 \, \text{cal/kg°C}) + (0.8 \, \text{kg} \cdot 0.0923 \, \text{cal/kg°C}) \right) \cdot 60°C \]

5. **Calculate Each Term:**
- For aluminum:
\[ Q_{Al} = 2.4 \cdot 0.216 = 0.5184 \, \text{cal} \]
- For brass:
\[ Q_{Brass} = 1.6 \cdot 0.0917 = 0.14672 \, \text{cal} \]
- For copper:
\[ Q_{Cu} = 0.8 \cdot 0.0923 = 0.07384 \, \text{cal} \]

6. **Sum the Contributions:**
\[ Q_{total} = 0.5184 + 0.14672 + 0.07384 = 0.73996 \, \text{cal} \]

7. **Multiply by ΔT:**
\[ Q_{net} = 0.73996 \cdot 60 = 44.3976 \, \text{cal} \]

8. **Final Result:**
Rounding to three significant figures, the total heat required is approximately:
\[ Q_{net} \approx 44.4 \, \text{cal} \]