Q1.

A cube of iron (density = 8000 kg/m³, specific heat capacity = 470 J/kg-K) is heated to a high temperature and is placed on a large block of ice at 0⁰C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temp of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg/m³ and the latent heat of fusion of ice = 3.36 * 10 ⁵ J/Kg.

Q2.

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. MOst of the energy needed for evaporation is taken from the water itself and the water is cooled down. ASsume that a pitcher contains 10 kg of water and 0.2 g water comes out per second. Assuming no backward reaction heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5⁰C. Specific heat capacity of water = 4200 J/Kg-⁰C and latent heat of vapourisation of water = 2.27 * 10⁶ J/kg.

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Let's tackle these two questions one at a time, starting with the first one about the iron cube and the ice. This problem involves concepts of heat transfer, specifically how the heat lost by the iron cube can melt the ice beneath it. We'll use the principles of specific heat and latent heat to find the initial temperature of the cube.

Understanding the Heat Transfer Process

When the hot iron cube is placed on the ice, it will lose heat, which will be used to melt the ice. The heat lost by the cube can be calculated using the formula:

Q = mcΔT

Where:

• Q = heat lost by the iron cube (in joules)
• m = mass of the iron cube (in kg)
• c = specific heat capacity of iron (in J/kg-K)
• ΔT = change in temperature (in K or °C)

The heat required to melt the ice can be calculated using:

Q = mL

Where:

• m = mass of ice melted (in kg)
• L = latent heat of fusion of ice (in J/kg)

Calculating the Mass of Ice Melted

First, we need to determine how much ice is melted by the cube. Since the cube sinks into the ice until its upper surface is just below the ice, we can assume that the volume of the cube equals the volume of the melted ice.

The density of ice is given as 900 kg/m³. The volume of the cube can be expressed as:

V = m_ice / ρ_ice

Where:

• m_ice = mass of ice melted
• ρ_ice = density of ice

Setting Up the Equations

Let’s denote the mass of the iron cube as m_iron. The heat lost by the iron cube can be expressed as:

Q_iron = m_iron * c_iron * (T_initial - 0)

Where T_initial is the initial temperature of the iron cube. The mass of the iron cube can be calculated from its density:

m_iron = V_iron * ρ_iron

Now, the heat gained by the ice can be expressed as:

Q_ice = m_ice * L_fusion

Equating Heat Lost and Gained

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

m_iron * c_iron * (T_initial - 0) = m_ice * L_fusion

Now, substituting the values:

• c_iron = 470 J/kg-K
• L_fusion = 3.36 * 10^5 J/kg
• ρ_iron = 8000 kg/m³
• ρ_ice = 900 kg/m³

Assuming the volume of the cube is V_iron = (m_iron / ρ_iron), we can express the mass of ice melted in terms of the volume of the cube:

m_ice = V_iron * ρ_ice = (m_iron / ρ_iron) * ρ_ice

Final Calculation

Substituting this back into our equation gives us:

m_iron * c_iron * T_initial = (m_iron / ρ_iron) * ρ_ice * L_fusion

Canceling m_iron from both sides, we can solve for T_initial:

T_initial = (ρ_ice * L_fusion) / (c_iron * ρ_iron)

Plugging in the values:

T_initial = (900 kg/m³ * 3.36 * 10^5 J/kg) / (470 J/kg-K * 8000 kg/m³)

Calculating this gives:

T_initial ≈ 79.7 °C

Cooling Water with a Porous Pitcher

Now, let's move on to the second question regarding the cooling of water in a porous pitcher. The process of evaporation cools the water, as it absorbs heat from the remaining water. We can calculate how long it takes for the temperature to decrease by 5 °C.

Understanding Evaporation and Heat Loss

When water evaporates, it takes away a significant amount of heat. The heat lost due to evaporation can be calculated using:

Q = m * L_vaporization

Where:

• m = mass of water evaporated (in kg)
• L_vaporization = latent heat of vaporization of water (in J/kg)

Calculating Heat Loss for Temperature Change

The heat required to decrease the temperature of the water can be calculated using:

Q = mcΔT

Where:

• m = mass of water (10 kg)
• c = specific heat capacity of water (4200 J/kg-K)
• ΔT = change in temperature (5 °C)

So, the heat required to cool the water by 5 °C is:

Q_cooling = 10 kg * 4200 J/kg-K * 5 K = 210000 J

Relating Heat Loss