An aluminium container of mass 100 gm contains 200 gm of ice at -20 C .Heat is added to the system at the rate of 100 cal/s . Find the temperature of the system after 4 minutes ( specific heat of ice = 0.5 and L= 80 cal/gm specific heat of Al= 0.2 cal/gm/C

To determine the final temperature of the system after adding heat for 4 minutes, we need to consider the heat transfer involved in warming the ice, melting it, and then warming the resulting water. We will break this down step by step, using the specific heat capacities and the latent heat of fusion for ice.

Initial Setup

We have an aluminum container and ice at a specific initial temperature. Here are the details:

• Mass of aluminum container (m_Al): 100 g
• Mass of ice (m_ice): 200 g
• Initial temperature of ice (T_ice): -20 °C
• Rate of heat addition: 100 cal/s
• Time of heat addition: 4 minutes (240 seconds)
• Specific heat of ice (c_ice): 0.5 cal/g°C
• Latent heat of fusion of ice (L): 80 cal/g
• Specific heat of aluminum (c_Al): 0.2 cal/g°C

Calculating Total Heat Added

The total heat added to the system over 4 minutes can be calculated as follows:

Q_total = Power × Time = 100 cal/s × 240 s = 24000 cal

Heating the Ice

First, we need to heat the ice from -20 °C to 0 °C. The heat required for this can be calculated using the formula:

Q_{heating ice} = m_ice × c_ice × ΔT

Where ΔT is the change in temperature (0 - (-20) = 20 °C).

Substituting the values:

Q_{heating ice} = 200 g × 0.5 cal/g°C × 20 °C = 2000 cal

Melting the Ice

Next, we need to calculate the heat required to melt the ice at 0 °C:

Q_{melting ice} = m_ice × L = 200 g × 80 cal/g = 16000 cal

Calculating Total Heat Required for Ice

The total heat required to first heat the ice to 0 °C and then melt it is:

Q_{total ice} = Q_{heating ice} + Q_{melting ice} = 2000 cal + 16000 cal = 18000 cal

Heat Available After Melting Ice

Now, we compare the total heat added to the heat required to completely melt the ice:

Since Q_total (24000 cal) > Q_{total ice} (18000 cal), we can melt all the ice and still have some heat left over.

Heat remaining after melting the ice:

Q_remaining = Q_total - Q_{total ice} = 24000 cal - 18000 cal = 6000 cal

Heating the Resulting Water

After melting, we have 200 g of water at 0 °C. Now, we will use the remaining heat to raise the temperature of this water. The specific heat of water is approximately 1 cal/g°C.

Using the formula:

Q_{heating water} = m_water × c_water × ΔT

We can rearrange this to find the change in temperature (ΔT):

ΔT = Q_remaining / (m_water × c_water)

Substituting the values:

ΔT = 6000 cal / (200 g × 1 cal/g°C) = 30 °C

Final Temperature Calculation

The final temperature of the system will be:

T_final = T_initial + ΔT = 0 °C + 30 °C = 30 °C

Thus, after 4 minutes of heating, the final temperature of the system is 30 °C.