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.
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]
1 answer