PLzzz Help MeeA cube of iron(density 8000kg/m^3,Specific heat capacity = 470J/kg-K)is heated to a high temprature and is placed on a large block of ice at 0 dEgree celsius. The cube melts the ice below it,displaces the water and sinks. IN the final equillibrium position,its upper surface just goes inside the ice. Calculate the initial temprature of cube.Neglect any loss of heat outside the ice and the cube. The density of ice = 900Kg/m^3 and latent heat of fusion of ICE = 3.36 x 10 ^5 J/Kg...

To determine the initial temperature of the iron cube, we need to analyze the heat transfer that occurs when the cube is placed on the ice. The heat lost by the iron cube will be equal to the heat gained by the ice as it melts. Let's break this down step by step.

Understanding the Problem

We have an iron cube with a density of 8000 kg/m³ and a specific heat capacity of 470 J/kg-K. When this cube is placed on ice at 0 degrees Celsius, it melts some of the ice, which then turns into water. The cube sinks until its upper surface is just below the ice. We need to find out the initial temperature of the cube before it was placed on the ice.

Key Values

• Density of iron: 8000 kg/m³
• Specific heat capacity of iron: 470 J/kg-K
• Density of ice: 900 kg/m³
• Latent heat of fusion of ice: 3.36 x 10^5 J/kg

Heat Transfer Calculation

When the iron cube is placed on the ice, it loses heat as it cools down to 0 degrees Celsius. This heat is used to melt the ice. The heat lost by the iron cube can be calculated using the formula:

Q_lost = m_iron * c_iron * (T_initial - T_final)

Where:

• Q_lost: Heat lost by the iron cube
• m_iron: Mass of the iron cube
• c_iron: Specific heat capacity of iron (470 J/kg-K)
• T_initial: Initial temperature of the iron cube
• T_final: Final temperature of the iron cube (0 degrees Celsius)

The mass of the iron cube can be expressed as:

m_iron = V_iron * density_iron

Where V_iron is the volume of the cube. If we assume the side length of the cube is a, then:

V_iron = a³

Thus, the mass becomes:

m_iron = a³ * 8000

Heat Gained by the Ice

The heat gained by the ice as it melts can be calculated using:

Q_gained = m_ice * L_fusion

Where:

• Q_gained: Heat gained by the ice
• m_ice: Mass of the ice melted
• L_fusion: Latent heat of fusion of ice (3.36 x 10^5 J/kg)

The mass of the ice melted can be calculated based on the volume of the water displaced by the iron cube:

m_ice = V_iron * density_ice

Substituting the volume of the iron cube, we have:

m_ice = a³ * 900

Setting Up the Equation

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

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

Substituting the expressions for mass:

(a³ * 8000) * 470 * T_initial = (a³ * 900) * (3.36 x 10^5)

Simplifying the Equation

We can cancel out a³ from both sides (assuming the cube is not zero volume):

8000 * 470 * T_initial = 900 * (3.36 x 10^5)

Calculating T_initial

Now, we can solve for T_initial:

T_initial = (900 * (3.36 x 10^5)) / (8000 * 470)

Calculating the right-hand side:

T_initial = (302400) / (3760000)

T_initial ≈ 0.0803 K

Final Result

The initial temperature of the iron cube is approximately 0.0803 K. This indicates that the cube was initially at a temperature just above 0 degrees Celsius, which is consistent with the scenario of it melting ice. In practical terms, this means the cube was likely at a temperature slightly above freezing, allowing it to melt the ice effectively.