Physics 2003 JAMB Past Questions - Question 26: if 1.2 x 10-J of heat energy is given off in 1 sec. from a vessel maintained at a temperature gradient of 30Km-¹ the surface area of the vessel is

Correct Answer

A

Explanation

To calculate the surface area of the vessel given the heat energy and temperature gradient, you can use the formula for heat conduction, which relates heat transfer to the thermal conductivity, surface area, temperature gradient, and thickness of the material. The formula is:

Q = (k * A * ΔT) / d

Where:
Q = Heat energy (1.2 x 10^-6 J)
k = Thermal conductivity of the material
A = Surface area of the vessel (we want to find this)
ΔT = Temperature gradient (30 K/m)
d = Thickness of the vessel (which is not provided, so we'll have to assume a value or work in terms of an infinitesimally thin vessel)

Since the thickness (d) is not given, you'll need to make an assumption about it. If you assume the vessel is infinitesimally thin (meaning heat transfer occurs across the surface without a distinct thickness), you can proceed to calculate the surface area.

Q = (k * A * ΔT) / d
1.2 x 10^-6 J = (k * A * 30 K/m) / (infinitesimally small thickness, which can be considered as close to zero)

In this case, the thickness (d) approaches zero, and the equation simplifies to:

1.2 x 10^-6 J = 0

This result doesn't make sense in the context of heat conduction, as you can't have zero thickness for the vessel. It's essential to know the thickness or make an assumption about it to calculate the surface area. If you have more information about the thickness or can provide more context, please let me know so I can assist you further.