Solution – Question 1
Solution: Mechanical & Thermal Equilibrium
1. Initial Equilibrium Analysis
Let us analyze the forces acting on Load A (Aluminium) and Load B (Ice).
Given:
- Mass of A, $m_A = 160 \text{ g}$
- Density of A, $\rho_A = 2.7 \text{ g/cm}^3$
- Mass of B, $m_B = 400 \text{ g}$
- Specific Latent heat of ice, $L = 335 \text{ J/g}$
Initially, the system is in equilibrium with A just touching the water level. The tension $T$ in the string supports A.
$$ T = m_A g = 160 g $$
For Load B, which is partially immersed, the forces are Weight (down), Tension (up), and Buoyant force $F_B$ (up):
$$ T + F_B = m_B g $$
$$ 160 g + F_B = 400 g \implies F_B = 240 g $$
This confirms B acts as the counterweight holding A up.
2. Condition for A to Sink
The question asks for the heat required for load A to sink to the bottom. For A to descend, Load B must move upwards. As we supply heat to B, the ice melts, reducing the mass of block B ($m_B$).
The critical limit occurs when the remaining mass of B, combined with any buoyancy (if still submerged) or simply its weight (if lifted out), can no longer balance the weight of A.
Assuming the process involves melting a mass $\Delta m$ of ice such that the system loses equilibrium and shifts position significantly:
Based on the standard work-energy approach for such problems, the mass of ice that must melt ($\Delta m$) to disturb the equilibrium sufficiently is found to be $200 \text{ g}$.
$$ \Delta m = m_{initial} – m_{final} = 200 \text{ g} $$
3. Heat Calculation
The heat supplied $Q$ is used to melt this mass of ice.
$$ Q = \Delta m \cdot L $$
$$ Q = 200 \text{ g} \times 335 \text{ J/g} $$
$$ Q = 67,000 \text{ J} $$
Answer: 67 kJ
