A circular pond of radius 12 m has a layer of ice 0.25 m thick. On a spring day, the temperature of the water below the ice is 4.0°C and the temperature of the air above is 11°C. What is the rate of heat transfer through the ice, in kW?
To find the rate of heat transfer through the ice, we can use the formula for heat transfer:
Q = kAΔT / d
where:
Q = rate of heat transfer
k = thermal conductivity of the ice
A = surface area of the pond
ΔT = temperature difference between the water and air
d = thickness of the ice layer
First, let's calculate the surface area of the pond:
A = πr²
A = π(12 m)²
A = π(144 m²)
A ≈ 452.39 m²
Next, let's calculate the temperature difference between the water and air:
ΔT = Ta - Tw
ΔT = 11°C - 4.0°C
ΔT = 7.0°C
Now, let's assume the thermal conductivity of ice, k, is approximately 2.2 W/(m·K). Convert this to kW/(m·°C):
k = 2.2 W/(m·K) * (1 kW/1000 W) * (1 K/°C)
k ≈ 0.0022 kW/(m·°C)
Finally, substitute these values into the formula for heat transfer:
Q = (0.0022 kW/(m·°C)) * (452.39 m²) * (7.0°C) / 0.25 m
Q = 0.0022 * 452.39 * 7.0 / 0.25 kW
Q ≈ 217.59 kW
Therefore, the rate of heat transfer through the ice is approximately 217.59 kW.
To find the rate of heat transfer through the ice, we can use the formula for heat conduction:
Q = k * A * ΔT / d
where:
Q is the rate of heat transfer,
k is the thermal conductivity of the material (ice in this case),
A is the surface area of the ice,
ΔT is the temperature difference between the water and air,
and d is the thickness of the ice.
Let's calculate the different components separately.
1. Surface Area of the Ice:
The surface area of the ice is equal to the circumference of the circular pond multiplied by the thickness of the ice.
Circumference = 2 * π * radius
Circumference = 2 * π * 12 m
Surface Area = Circumference * thickness
Surface Area = (2 * π * 12 m) * (0.25 m)
2. Temperature Difference:
The temperature difference between the water and air is the difference between the temperature of the water below the ice and the temperature of the air above.
ΔT = Temperature of water - Temperature of air
ΔT = 4.0°C - 11.0°C
3. Thermal Conductivity of Ice:
The thermal conductivity of ice depends on its temperature. At 0°C, the thermal conductivity of ice is approximately 2.2 W/(m·K). However, since the water below the ice is slightly warmer, we can assume a slightly higher thermal conductivity. Let's use a value of 2.5 W/(m·K) for simplicity.
Now, let's substitute these values into the formula:
Q = k * A * ΔT / d
Q = (2.5 W/(m·K)) * (2 * π * 12 m * 0.25 m) * (4.0°C - 11.0°C)
The temperature difference, 4.0°C - 11.0°C, is equal to -7.0°C.
Q = 2.5 * (2 * π * 12 * 0.25) * -7.0
Simplifying further:
Q = -2.5 * (2 * 3.14 * 12 * 0.25) * 7
Calculating this expression gives us the rate of heat transfer through the ice in watts. To convert watts to kilowatts (kW), we divide the result by 1000.
Finally, calculate the value to find the rate of heat transfer through the ice in kW.