An ice cube of mass 0.1Kg is placed in an isolated container which is at 227^oc .The specific heat S of the container varies with temperature T according to the empirical relation. S=A+BT,whereA=100cal/KgK and B=2*10² cal /KgK².If the final temperature of the container is 27⁰C.Determine the mass of the container.[Latent heat of fusion of water=8*10⁴cal/Kg,specific of heat of water=10³cal/Kg]

To solve the problem of determining the mass of the container, we need to apply the principles of heat transfer, specifically focusing on the heat lost by the container and the heat gained by the ice cube. Let's break this down step by step.

Understanding the Heat Transfer

When the ice cube is placed in the container, it will absorb heat from the container until it reaches thermal equilibrium at the final temperature of 27°C. The heat lost by the container will equal the heat gained by the ice cube, which includes the heat required to melt the ice and then raise the temperature of the resulting water to 27°C.

Key Variables and Constants

• Mass of ice cube (m_ice): 0.1 kg
• Initial temperature of the container (T_container_initial): 227°C
• Final temperature (T_final): 27°C
• Latent heat of fusion of water (L_f): 8 x 10⁴ cal/kg
• Specific heat of water (c_water): 10³ cal/kgK
• Specific heat of the container (S): S = A + BT, where A = 100 cal/kgK and B = 2 x 10² cal/kgK²

Calculating Heat Gained by the Ice Cube

The ice cube will first need to melt, and then the resulting water will be heated to 27°C. The total heat gained by the ice cube (Q_ice) can be calculated as follows:

• Heat to melt the ice: Q_melt = m_ice * L_f = 0.1 kg * 8 x 10⁴ cal/kg = 8 x 10³ cal
• Heat to raise the temperature of the melted ice (water) to 27°C: Q_heat = m_ice * c_water * ΔT = 0.1 kg * 10³ cal/kgK * (27°C - 0°C) = 0.1 * 10³ * 27 = 2.7 x 10⁴ cal

Thus, the total heat gained by the ice cube is:

Q_ice = Q_melt + Q_heat = 8 x 10³ cal + 2.7 x 10⁴ cal = 3.5 x 10⁴ cal

Calculating Heat Lost by the Container

The heat lost by the container (Q_container) can be expressed as:

Q_container = m_container * S * ΔT

Where:

• m_container: mass of the container (unknown)
• ΔT: change in temperature of the container = (227°C - 27°C) = 200°C

To find the specific heat of the container at the initial temperature of 227°C, we substitute T into the empirical relation:

S = A + BT = 100 cal/kgK + (2 x 10² cal/kgK² * 227) = 100 + 45400 = 45500 cal/kgK

Now we can express the heat lost by the container:

Q_container = m_container * 45500 cal/kgK * 200 K

Q_container = m_container * 9100000 cal

Setting Up the Equation

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

m_container * 9100000 cal = 3.5 x 10⁴ cal

Solving for the Mass of the Container

Now, we can solve for the mass of the container:

m_container = (3.5 x 10⁴ cal) / (9100000 cal/kg)

m_container ≈ 0.00385 kg

Final Result

The mass of the container is approximately 0.00385 kg, or 3.85 grams. This calculation illustrates how energy conservation principles apply in thermal systems, where heat lost by one body equals heat gained by another, allowing us to find unknown quantities in thermodynamic processes.

To solve this problem, we need to analyze the heat transfer between the ice cube and the container. The ice cube will absorb heat from the container, causing it to melt and then warm up to the final equilibrium temperature. We will use the principles of thermodynamics, specifically the concepts of specific heat and latent heat, to find the mass of the container.

Understanding the Heat Transfer Process

First, let's break down the components involved:

• The mass of the ice cube, m_ice = 0.1 kg.
• The initial temperature of the container, T_container_initial = 227 °C.
• The final temperature of both the ice and the container, T_final = 27 °C.
• The latent heat of fusion of ice, L_f = 8 × 10^4 cal/kg.
• The specific heat of water, c_water = 10^3 cal/kg·K.
• The specific heat of the container varies with temperature as S = A + BT, where A = 100 cal/kg·K and B = 2 × 10^2 cal/kg·K².

Calculating the Heat Absorbed by the Ice Cube

When the ice cube is placed in the container, it first absorbs heat to melt, and then the resulting water warms up to the final temperature. The total heat absorbed by the ice cube can be calculated in two parts:

1. Heat required to melt the ice:

Q_melt = m_ice × L_f = 0.1 kg × 8 × 10^4 cal/kg = 8 × 10^3 cal.

2. Heat required to warm the melted ice (water) from 0 °C to 27 °C:

Q_warm = m_ice × c_water × ΔT = 0.1 kg × 10^3 cal/kg·K × (27 °C - 0 °C) = 0.1 kg × 10^3 cal/kg·K × 27 K = 2.7 × 10^3 cal.

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

Q_total_ice = Q_melt + Q_warm = 8 × 10^3 cal + 2.7 × 10^3 cal = 10.7 × 10^3 cal.

Calculating the Heat Lost by the Container

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

The heat lost by the container can be expressed as:

Q_lost = m_container × ∫(S(T) dT)

Where:

S(T) = A + BT = 100 + 200T

We need to integrate this from the initial temperature (227 °C) to the final temperature (27 °C):

Q_lost = m_container × ∫(100 + 200T) dT from 227 to 27.

Calculating the integral:

∫(100 + 200T) dT = 100T + 100T²

Evaluating this from 227 to 27:

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

Calculating the values:

Q_lost = m_container × [2700 + 100(729) - (22700 + 100(51529))]

Q_lost = m_container × [2700 + 72900 - 22700 - 5152900]

Q_lost = m_container × [72900 - 5152900]

Q_lost = m_container × (-5152900 + 72900)

Q_lost = m_container × (-5080000)

Setting Heat Lost Equal to Heat Gained

Since the heat lost by the container equals the heat gained by the ice cube, we can set these two equations equal to each other:

10.7 × 10^3 cal = m_container × (-5080000)

Now, solving for the mass of the container:

m_container = 10.7 × 10^3 cal / 5080000 cal/kg

m_container ≈ 0.0021 kg

Final Result

The mass of the container is approximately 0.0021 kg, or 2.1 grams. This calculation illustrates the principles of heat transfer and the relationship between mass, specific heat, and temperature change in a thermodynamic system.