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Step-by-Step Solution
Step 1: List the given data
• Mass of aluminium sphere, $m_{\text{Al}} = 0.20 \text{ kg}$
• Initial temperature of aluminium, $T_{\text{Al, initial}} = 150^\circ \text{C}$
• Volume of water, $V_{\text{water}} = 150 \text{ cc} = 150 \text{ mL} \approx 0.15 \text{ kg}$ (since $1 \text{ mL}$ of water $\approx 1 \text{ g}$)
• Initial temperature of water, $T_{\text{water, initial}} = 27^\circ \text{C}$
• Water equivalent of the calorimeter, $m_{\text{cal}} = 0.025 \text{ kg}$
• Final temperature of the system, $T_{\text{final}} = 40^\circ \text{C}$
• Let the specific heat of aluminium be $S$ (in $ \text{J} \, \text{kg}^{-1} \, {}^\circ\text{C}^{-1}$).
• Specific heat of water, $c_{\text{water}} = 4200 \,\text{J} \,\text{kg}^{-1}\,{}^\circ\text{C}^{-1}$ (or $1 \,\text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$ but often taken as $4200 \,\text{J} \,\text{kg}^{-1}\,{}^\circ\text{C}^{-1}$).
Step 2: State the principle of calorimetry
When two bodies at different temperatures are brought into contact in a calorimeter (assumed perfectly insulated), the heat lost by the hotter body equals the heat gained by the cooler bodies. Symbolically:
$Q_{\text{lost}} = Q_{\text{gained}}.$
Step 3: Apply the principle of calorimetry
• Heat lost by the aluminium sphere when it cools from $150^\circ \text{C}$ to $40^\circ \text{C}$ is:
$Q_{\text{Al}} = m_{\text{Al}} \times S \times \bigl(T_{\text{Al, initial}} - T_{\text{final}}\bigr).$
• Heat gained by the water when it warms from $27^\circ \text{C}$ to $40^\circ \text{C}$ is:
$Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \bigl(T_{\text{final}} - T_{\text{water, initial}}\bigr).$
• Heat gained by the calorimeter (treated as equivalent mass of water) is:
$Q_{\text{cal}} = m_{\text{cal}} \times c_{\text{water}} \times \bigl(T_{\text{final}} - T_{\text{water, initial}}\bigr).$
Step 4: Write the heat balance equation
According to the principle of calorimetry:
$m_{\text{Al}} \times S \times \bigl(150 - 40\bigr) = m_{\text{water}} \times c_{\text{water}} \times \bigl(40 - 27\bigr) + m_{\text{cal}} \times c_{\text{water}} \times \bigl(40 - 27\bigr).$
Substituting the values:
$0.20 \times S \times (110) = 0.15 \times 4200 \times (13) + 0.025 \times 4200 \times (13).$
Step 5: Compute the right-hand side
Calculate the heat gained by water plus calorimeter:
$Q_{\text{water}} = 0.15 \times 4200 \times 13.$
$Q_{\text{cal}} = 0.025 \times 4200 \times 13.$
Let’s compute each term separately (though typically you can combine them):
$Q_{\text{water}} = 0.15 \times 4200 \times 13 = 0.15 \times 4200 \times 13 = 0.15 \times 54600 = 8190 \,\text{J}.$
$Q_{\text{cal}} = 0.025 \times 4200 \times 13 = 0.025 \times 54600 = 1365 \,\text{J}.$
Hence, total heat gained:
$Q_{\text{water}} + Q_{\text{cal}} = 8190 + 1365 = 9555 \,\text{J}.$
Step 6: Solve for the specific heat $S$
The left-hand side is:
$0.20 \times S \times 110 = 9555.$
So,
$S = \frac{9555}{0.20 \times 110} = \frac{9555}{22} = 434 \,\text{J} \,\text{kg}^{-1}\,{}^\circ\text{C}^{-1}.$
Step 7: Conclusion
Therefore, the specific heat of aluminium is $434 \,\text{J} \,\text{kg}^{-1}\,{}^\circ\text{C}^{-1}.$
