Question

In an experiment a sphere of aluminium of mass 0.20 kg is heated upto $150^{\circ}C$. Immediately, it is put into water of volume 150 cc at $27^{\circ}C$ kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is $40^{\circ}C$. The specific heat of aluminium is : (take 4.2 Joule=1 calorie)

• A. $378 \, J/kg - {^{\circ}C}$
• B. $315 \, J/kg - {^{\circ}C}$
• C. $476 \, J/kg - {^{\circ}C}$
• D. $434 \, J/kg - {^{\circ}C}$

Answer 1

The Correct Option is D

To find the specific heat of aluminium, we need to analyze the heat transfer between the aluminium sphere, water, and the calorimeter. The principle to use here is the law of conservation of energy, where the heat lost by the aluminium sphere is equal to the heat gained by the water and the calorimeter.

1. Identify the masses and specific heats:
   • Mass of aluminium sphere, m_{Al} = 0.20 \, \text{kg}
   • Initial temperature of aluminium, T_{Al\_initial} = 150^{\circ}C
   • Mass of water, m_{water} = 150 \, \text{cc} = 0.150 \, \text{kg} (since density of water is 1 \, \text{g/cc})
   • Initial temperature of water, T_{water\_initial} = 27^{\circ}C
   • Water equivalent of calorimeter, m_{cal} = 0.025 \, \text{kg}
   • Final temperature of the entire system, T_{final} = 40^{\circ}C
   • Specific heat of water, c_{water} = 1 \, \text{cal/g}^{\circ}C = 4.2 \, \text{J/g}^{\circ}C = 4200 \, \text{J/kg}^{\circ}C
2. Calculate the heat lost by the aluminium:
   • The formula for heat lost is Q = m \cdot c \cdot \Delta T, where \Delta T is the change in temperature.
   • For the aluminium sphere: Q_{Al} = m_{Al} \cdot c_{Al} \cdot (T_{Al\_initial} - T_{final})
   • Substituting the values: Q_{Al} = 0.20 \cdot c_{Al} \cdot (150 - 40) = 0.20 \cdot c_{Al} \cdot 110
   • Thus, Q_{Al} = 22 \cdot c_{Al}
3. Calculate the heat gained by the water and calorimeter:
   • For water: Q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{water\_initial})
   • Substitute the values: Q_{water} = 0.150 \times 4200 \times (40 - 27) = 8190 \, \text{J}
   • For the calorimeter: Q_{cal} = m_{cal} \cdot c_{water} \cdot (T_{final} - T_{water\_initial})
   • Substitute the values: Q_{cal} = 0.025 \times 4200 \times (40 - 27) = 1365 \, \text{J}
4. Total heat gained by the water and calorimeter:
   • Total Q_g = Q_{water} + Q_{cal} = 8190 + 1365 = 9555 \, \text{J}
5. According to the conservation of energy, the heat lost by aluminium equals the heat gained by water and calorimeter:
   • 22 \cdot c_{Al} = 9555
   • Solving for c_{Al}, we get c_{Al} = \frac{9555}{22} \approx 434 \, \text{J/kg}^{\circ}C

The specific heat of aluminium is 434 \, J/kg - {^{\circ}C}.