Let us consider the problem of heat transfer through the surface of ice. Let S = 1 m² be the surface area of the ice and h = 25 cm be the thickness of the ice.
The air temperature above the ice surface is t1 = 20°C, and the water temperature at the ice surface is t2 = 0°C. It is necessary to determine the amount of heat that will pass through the surface of the ice in 1 hour.
To solve the problem, we use the formula for calculating heat transfer through a flat layer:
Q = k * S * (t1 - t2) / h
where Q is the amount of heat that will pass through the layer in 1 hour, k is the thermal conductivity coefficient of the material, S is the surface area, t1 and t2 are the temperatures on one and the other side of the layer, respectively, h is the thickness of the layer.
For ice, the thermal conductivity coefficient k = 2.22 W/(m K).
Substituting known values, we get:
Q = 2.22 * 1 * (20 - 0) / 0.25 = 355.2 W.
Consequently, 355.2 W of heat will pass through the ice surface in 1 hour.
To answer the second question, consider the intensity of heat transfer through air and ice in the absence of heat transfer by convection and radiation.
The intensity of heat transfer through air is determined by the formula:
Qв = α * S * (t1 - t2)
where α is the heat transfer coefficient, depending on the air speed, S is the surface area, t1 and t2 are the temperatures on one and the other side of the surface, respectively.
For air at rest α ≈ 10 W/(m²·K).
Substituting known values, we get:
Qв = 10 * 1 * (20 - 0) = 200 W.
Thus, the intensity of heat transfer through ice is approximately 1.8 times higher (355.2 W / 200 W) compared to heat transfer through air in the absence of heat transfer by convection and radiation.
Our digital product is a convenient calculator for calculating the amount of heat that will pass in 1 hour through an ice surface with a thickness of 25 cm and an area of 1 m² at given air and water temperatures. This tool will be useful for people associated with ice facilities and structures such as cold storage rooms, skating rinks and ice arenas.
With our calculator you can quickly calculate the amount of heat that will pass through the surface of the ice in 1 hour using the formula for calculating heat transfer through a flat layer. Also in our calculator you can find out how many times the intensity of heat transfer through ice is higher than through air, in the absence of heat transfer by convection and radiation.
Our calculator is easy to use and can be used by both experts and amateurs. Simply enter the air and water temperatures, as well as the ice surface area, and click on the "Calculate" button. The results will be displayed on the screen within a few seconds.
The task is to determine the amount of heat that will pass through the surface of the ice in 1 hour under given conditions. It is also necessary to determine how many times the intensity of heat transfer through ice is higher than through air, in the absence of heat transfer by convection and radiation.
From the problem conditions it is known that the ice surface area is S = 1 m², the ice thickness is h = 25 cm, the air temperature above the ice surface is t1 = 20°C, and the water temperature near the ice surface is t2 = 0°C. Thermal conductivity coefficient of ice k = 2.22 W/(m K).
To solve the problem, we use the formula for calculating heat transfer through a flat layer: Q = k * S * (t1 - t2) / h
where Q is the amount of heat that will pass through the layer in 1 hour, k is the thermal conductivity coefficient of the material, S is the surface area, t1 and t2 are the temperatures on one and the other side of the layer, respectively, h is the thickness of the layer.
Substituting known values, we get: Q = 2.22 * 1 * (20 - 0) / 0.25 = 355.2 W
Consequently, 355.2 W of heat will pass through the ice surface in 1 hour.
To determine how many times the intensity of heat transfer through ice is higher than through air, in the absence of heat transfer by convection and radiation, we use the formula for calculating the intensity of heat transfer through air: Qв = α * S * (t1 - t2)
where α is the heat transfer coefficient, depending on the air speed, S is the surface area, t1 and t2 are the temperatures on one and the other side of the surface, respectively.
For air at rest α ≈ 10 W/(m²·K).
Substituting known values, we get: Qv = 10 * 1 * (20 - 0) = 200 W
Thus, the intensity of heat transfer through ice is approximately 1.8 times higher (355.2 W / 200 W) compared to heat transfer through air in the absence of heat transfer by convection and radiation.
Thus, the amount of heat that will pass in 1 hour through the ice surface S=1m² with thickness h=25cm, if the air temperature is t1=20°C, and the water temperature at the ice surface is t2=0°C, is equal to 355.2 W, and the intensity of heat transfer through ice is approximately 1.8 times higher compared to heat transfer through air in the absence of heat transfer by convection and radiation.
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To solve this problem it is necessary to use Fourier's law of thermal conductivity.
From the problem conditions it is known that the temperature on one side of the ice surface is 20°C (air temperature), and on the other side it is 0°C (water temperature at the ice surface). It is also known that the thickness of the ice is 25 cm (that is, h = 0.25 m).
The intensity of heat transfer through ice can be calculated using the formula:
Q = k * S * ΔT / d,
where Q is the amount of heat passing through the surface per unit time (in this case, 1 hour), k is the thermal conductivity coefficient of ice, S is the surface area (in this case S = 1 m^2), ΔT is the temperature difference between the ice surfaces (20°C - 0°C = 20°C), d - ice thickness (0.25 m).
The value of the thermal conductivity coefficient of ice can be found in tables of physical properties of substances. For ice at a temperature of 0°C, the thermal conductivity coefficient k ≈ 2.2 W/(m K).
Thus, substituting known values into the formula, we get:
Q = 2.2 W/(m K) * 1 m^2 * 20°C / 0.25 m * 3600 s = 63360 W = 63.36 kW.
Answer: 63.36 kW of heat will pass through an ice surface 25 cm thick in 1 hour.
To answer the second part of the question, it is necessary to calculate the intensity of heat transfer through the air. To do this, you can use a similar formula:
Q = k * S * ΔT / d,
where k is the thermal conductivity coefficient of air, which is orders of magnitude lower than that of ice, S and d are the same as before, ΔT is the temperature difference between the surfaces (20°C - 0°C = 20°C).
The thermal conductivity coefficient of air at room temperature (20°C) is k ≈ 0.026 W/(m·K).
Substituting the known values into the formula, we get:
Q = 0.026 W/(m K) * 1 m^2 * 20°C / 0.25 m * 3600 s = 936.96 W = 0.94 kW.
Thus, the intensity of heat transfer through ice is approximately 67 times higher than through air (63.36 kW / 0.94 kW ≈ 67).
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