43. If 1.2 x 106
J of heat energy is
given off in 1sec from a vessel
maintained at a temperature
gradient of 30km-1, the surface
area of the vessel is___
A. 1.0 x 103m2
B. 1.0 x 102m2
C. 9.0 x 104m2
D. 9.0 x 102m2
[Thermal capacity of the vessel =
400Wm-1K
-1]
To find the surface area of the vessel, we can use the formula:
Q = k * A * dT
Where Q is the heat energy given off in 1 second (1.2 x 10^6 J), k is the thermal capacity of the vessel (400 Wm^-1K^-1), A is the surface area of the vessel, and dT is the temperature gradient (30 km^-1).
Rearranging the formula, we have:
A = Q / (k * dT)
Substituting the given values, we have:
A = (1.2 x 10^6 J) / (400 Wm^-1K^-1 * 30 km^-1)
Converting km to m, we get:
A = (1.2 x 10^6 J) / (400 Wm^-1K^-1 * 30,000 m^-1)
Simplifying, we have:
A = 1.2 x 10^6 / (12 x 10^4 * 3 x 10^4) m^2
A = 1.2 x 10^6 / (3.6 x 10^8) m^2
A = 1.2 / 3.6 x 10^-2 m^2
A = 1 / 3 x 10^-2 m^2
A = 10^2 / 3 m^2
A = 100/3 m^2
A ≈ 33.33 m^2
Therefore, the surface area of the vessel is approximately 33.33 m^2.
None of the given options match this result.
To find the surface area of the vessel, first, let's use the formula for the heat transfer rate:
Q/t = k * A * ΔT / L
Where:
Q = Heat energy given off (1.2 x 10^6 J)
t = Time (1 sec)
k = Thermal capacity of the vessel (400 W/m^-1K^-1)
A = Surface area of the vessel (what we need to find)
ΔT = Temperature gradient (30 km^-1)
L = Thickness of the vessel
Rearranging the equation to solve for A:
A = (Q * L) / (k * ΔT * t)
Substituting the given values:
A = (1.2 x 10^6 J * 1 sec) / (400 W/m^-1K^-1 * 30 km^-1 * 1 sec)
Converting km to meters (1 km = 1000 m):
A = (1.2 x 10^6 J) / (400 W/m^-1K^-1 * 30 * 1000 m^-1)
Simplifying:
A = 1.0 x 10^6 J / (12,000 W/m^-1K^-1)
A = 1.0 x 10^6 J / 12,000 W
A = 83.33 m^2
Therefore, the surface area of the vessel is 83.33 m^2. However, none of the given options (A, B, C, or D) match this result. Please double-check the options provided or check if there was any mistake in the given information.