An ice cube of mass 0·1 Kg. at 0º C is placed in an isolated container which is at 227ºC. The specific heat S of the
container varies with temperature T according to the empirical relation S = A + BT, where A = 100 cal / Kg – K and
B = 2 × 10^-2 cal / Kg – K2. If the final temperature of the container is 27ºC, determine the mass of the container.
(Latent heat of fusion for water=8×10^4 cal / Kg. specific heat of water = 10^3 cal /Kg – K).

To solve this problem, we need to analyze the heat transfer between the ice cube and the container. The ice cube will absorb heat to melt and then warm up to the final temperature of the system, while the container will lose heat as it cools down. Let's break this down step by step.

Understanding the Heat Transfer Process

The process involves two main stages for the ice cube:

• Melting the ice cube at 0°C to water at 0°C.
• Heating the resulting water from 0°C to the final temperature of 27°C.

For the container, it will lose heat as its temperature decreases from 227°C to 27°C. We will use the specific heat formula and the latent heat of fusion to calculate the heat involved in each step.

Calculating Heat Absorbed by the Ice Cube

First, we need to calculate the heat required to melt the ice cube:

The formula for heat absorbed during melting is:

Q_melt = m_ice * L_f

Where:

• m_ice = mass of the ice cube = 0.1 kg
• L_f = latent heat of fusion = 8 × 10⁴ cal/kg

Substituting the values:

Q_melt = 0.1 kg * 8 × 10⁴ cal/kg = 8000 cal

Next, we calculate the heat required to raise the temperature of the melted water from 0°C to 27°C:

Q_heat = m_water * c_water * ΔT

Where:

• m_water = 0.1 kg
• c_water = specific heat of water = 10³ cal/kg·K
• ΔT = change in temperature = 27°C - 0°C = 27 K

Substituting the values:

Q_heat = 0.1 kg * 10³ cal/kg·K * 27 K = 2700 cal

Now, we can find the total heat absorbed by the ice cube:

Q_{total, ice} = Q_melt + Q_heat = 8000 cal + 2700 cal = 10700 cal

Calculating Heat Lost by the Container

Next, we need to calculate the heat lost by the container as it cools down from 227°C to 27°C. The specific heat of the container varies with temperature, so we will integrate the specific heat over the temperature range:

Let m_container be the mass of the container. The heat lost by the container can be expressed as:

Q_lost = m_container * ∫(S(T) dT)

Where S(T) = A + BT = 100 + 0.02T. The limits of integration will be from 27°C to 227°C:

Calculating the integral:

∫(100 + 0.02T) dT = 100T + 0.01T² + C

Evaluating from 27 to 227:

Q_lost = m_container * [100(227) + 0.01(227)² - (100(27) + 0.01(27)²)]

Calculating the individual terms:

100(227) = 22700

0.01(227)² = 0.01 * 51529 = 515.29

100(27) = 2700

0.01(27)² = 0.01 * 729 = 7.29

Now substituting these values:

Q_lost = m_container * [22700 + 515.29 - (2700 + 7.29)]

Q_lost = m_container * [22700 + 515.29 - 2700 - 7.29]

Q_lost = m_container * [20008] cal

Setting Up the Equation

Since the heat gained by the ice cube equals the heat lost by the container, we can set up the equation:

10700 cal = m_container * 20008 cal

Now, solving for m_container:

m_container = 10700 cal / 20008 cal/kg

m_container ≈ 0.534 kg

Final Result

The mass of the container is approximately 0.534 kg. This calculation shows how energy conservation principles apply in thermal systems, where heat lost by one body is equal to the heat gained by another, leading to a final equilibrium temperature.

Approved