The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.5 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 37 ˚C and that of the cooler surface is 7 ˚C Determine the heat that flows in 7 s for each case.
To determine the heat that flows in 7 seconds for each case, we can use the formula for heat transfer:
Q = kAΔT/t
Where:
Q = Heat transferred (in Joules)
k = Thermal conductivity (in J/(s·m·C˚))
A = Cross-sectional area (in m²)
ΔT = Temperature difference (in ˚C)
t = Time (in seconds)
Let's calculate the heat flow for each case.
Case A:
In Case A, heat is conducted in the x-direction. The cross-sectional area of the block in the x-direction is 2L0 × 3L0 = 6L0².
ΔT = 37 ˚C - 7 ˚C = 30 ˚C
Plugging in the values:
Q = (150 J/(s·m·C˚)) * (6L0²) * (30 ˚C) / 7 s
Now, we need to substitute the value of L0 (0.5 m) into the equation:
Q = (150 J/(s·m·C˚)) * (6(0.5 m)²) * (30 ˚C) / 7 s
Simplifying:
Q = (150 J/(s·m·C˚)) * (6 * 0.25 m²) * (30 ˚C) / 7 s
Q = (150 J/(s·m·C˚)) * (1.5 m²) * (30 ˚C) / 7 s
Q = 150 * 1.5 * 30 / 7 J
So, the heat flow in Case A in 7 seconds is approximately 321.43 Joules.
Now, let's calculate the heat flow for Case B and Case C using the same formula and the given information.
To determine the heat that flows in 7 seconds for each case, we can use the formula for heat conduction:
Q = (k * A * ΔT) / L
Where:
Q is the heat transferred (in Joules)
k is the thermal conductivity of the block (in J/(s·m·C˚))
A is the cross-sectional area of heat flow (in m^2)
ΔT is the temperature difference between the warmer and cooler surfaces (in ˚C)
L is the thickness of the block (in meters)
Let's calculate the heat flow for each case:
Case A: Heat flows in the x-direction
The cross-sectional area (A) is L0 * 2L0 = 1m * 2m = 2m^2
The thickness (L) is 3L0 = 3 * 0.5m = 1.5m
The temperature difference (ΔT) is 37˚C - 7˚C = 30˚C
Plugging these values into the formula:
Q = (150 J/(s·m·C˚) * 2m^2 * 30˚C) / 1.5m
Q = (9000 J/˚C) / 1.5
Q = 6000 J
Therefore, the heat flow in 7 seconds for Case A is 6000 Joules.
Case B: Heat flows in the y-direction
The cross-sectional area (A) is L0 * 3L0= 1m * 3m = 3m^2
The thickness (L) is 2L0 = 2 * 0.5m = 1m
The temperature difference (ΔT) is 37˚C - 7˚C = 30˚C
Plugging these values into the formula:
Q = (150 J/(s·m·C˚) * 3m^2 * 30˚C) / 1m
Q = (13500 J/˚C) / 1
Q = 13500 J
Therefore, the heat flow in 7 seconds for Case B is 13500 Joules.
Case C: Heat flows in the z-direction
The cross-sectional area (A) is 2L0 * 3L0 = 2m * 3m = 6m^2
The thickness (L) is L0 = 0.5m
The temperature difference (ΔT) is 37˚C - 7˚C = 30˚C
Plugging these values into the formula:
Q = (150 J/(s·m·C˚) * 6m^2 * 30˚C) / 0.5m
Q = (27000 J/˚C) / 0.5
Q = 54000 J
Therefore, the heat flow in 7 seconds for Case C is 54000 Joules.
In summary, the heat flows in 7 seconds are as follows:
- Case A: 6000 Joules
- Case B: 13500 Joules
- Case C: 54000 Joules