Physics 2003 JAMB Past Questions - Question 25: A 2000W electric heater is used to heat a metal object of mass 5kg initially at 10°C. If a temperature rise of 30°C is obtained after 10min, the heat capacity of the material is

Correct Answer

D

Explanation

To find the heat capacity of the material, you can use the formula:

Q = m * C * ΔT

Where:
Q = heat energy (in joules)
m = mass of the material (in kilograms)
C = specific heat capacity of the material (in J/(kg·°C))
ΔT = change in temperature (in °C)

You are given the following information:
- Power of the electric heater (P) = 2000 watts = 2000 joules per second (1 watt = 1 joule/second).
- Time (t) = 10 minutes = 600 seconds.
- Initial temperature (T1) = 10°C.
- Temperature rise (ΔT) = 30°C.
- Mass (m) = 5 kg.

First, let's find the total energy (Q) supplied by the electric heater during the 10 minutes:

Q = P * t
Q = 2000 J/s * 600 s
Q = 1,200,000 joules

Now, we can calculate the specific heat capacity (C) of the material using the formula:

Q = m * C * ΔT

1,200,000 J = 5 kg * C * 30°C

Now, solve for C:

C = 1,200,000 J / (5 kg * 30°C)
C = 8,000 J/(kg·°C)

So, the specific heat capacity of the material is 8,000 J/(kg· ° C).