To find the initial temperature of the metal ball, we will equate the heat lost by the ball to the heat gained by the ice during melting.

Given:

• Specific gravity of the metal ball $S_b = 4.5$
• Specific heat of the metal $c = 0.1 \, \text{cal/gm–°C}$
• Latent heat of ice $L = 80 \, \text{cal/g}$
• Specific gravity of ice $S_i = 0.9$
• Final temperature of the ball $T_f = 0^\circ \text{C}$
• Initial temperature of the ball $T_i$ (to be found)

Step 1: Mass of the Ball

Since the specific gravity is the ratio of the density of the object to the density of water: $\text{Density of metal ball} = S_b \times \text{Density of water}$ $\text{Density of metal ball} = 4.5 \times 1 \, \text{gm/cm}^3 = 4.5 \, \text{gm/cm}^3$ Let the volume of the ball be $V$. Therefore, the mass of the ball $M$ is: $M = 4.5 \times V \, \text{gm}$

Step 2: Mass of Ice Melted

When the ball reaches 0°C, half of it sinks into the ice. For half of the ball to sink, the weight of the displaced ice must be equal to half of the weight of the ball.

The volume of ice displaced $V_i$ is given by: $\text{Mass of displaced ice} = V_i \times \text{Density of ice} = V_i \times 0.9 \, \text{gm/cm}^3$ Since the weight of displaced ice equals half the weight of the ball: $0.9 \times V_i = \frac{1}{2} \times 4.5 \times V$ Solving for $V_i$: $V_i = \frac{4.5 \times V}{2 \times 0.9} = \frac{4.5V}{1.8} = 2.5V$ So the mass of ice melted $m_i$ is: $m_i = 0.9 \times V_i = 0.9 \times 2.5V = 2.25V \, \text{gm}$

Step 3: Heat Gained by Ice

The heat gained by the ice during melting is: $Q_{\text{gained}} = m_i \times L = 2.25V \times 80 \, \text{cal} = 180V \, \text{cal}$

Step 4: Heat Lost by Ball

The heat lost by the ball as it cools from $T_i$ to $0^\circ \text{C}$ is: $Q_{\text{lost}} = M \times c \times (T_i - 0) = 4.5V \times 0.1 \times T_i = 0.45V \times T_i \, \text{cal}$

Step 5: Equating Heat Lost and Gained

By the principle of conservation of energy: $Q_{\text{lost}} = Q_{\text{gained}}$ $0.45V \times T_i = 180V$ Cancel out $V$ on both sides: $0.45 \times T_i = 180$ Solving for $T_i$: $T_i = \frac{180}{0.45} = 400^\circ \text{C}$

So, the initial temperature of the metal ball is $\boxed{400^\circ \text{C}}$.

Do you want more details on any step or have any questions?

Related Questions:

1. What would happen if the specific gravity of the ball were different?
2. How would the problem change if the specific heat of the ball were doubled?
3. What if the latent heat of ice was different, how would that affect the result?
4. How is the specific gravity of a substance related to its density?
5. Could the initial temperature be determined if we only knew the heat gained by the ice?

Tip:

Always ensure the units of all physical quantities match when solving thermodynamics problems, as this can prevent errors in calculations.