Calculate the specific heat capacity of a metal rod of mass 0.025kg whose temperature was raised by 15ºC when 1000J of heat energy was added to the rod(assuming the heat loss to the surrounding is negligible)
A
2440.5JKg\(^{-1}\)K\(^{-1}\)
B
3000.0JKg\(^{-1}\)K\(^{-1}\)
C
2888.4JKg\(^{-1}\)K\(^{-1}\)
D
2666.7JKg\(^{-1}\)K\(^{-1}\)
Verified Correct Answer: D.
Physical Principles & Numerical Solution
D. 2666.7 \(\text{J kg}^{-1} \, \text{K}^{-1}\)
Using the formula: \(Q = mc\Delta T\)
We can rearrange to find \( c \): \(c = \frac{Q}{m \Delta T} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\)
Thus, the specific heat capacity is:
\(c = 2666.7 \, \text{J kg}^{-1} \, \text{K}^{-1}\).
Source: Myschool Past Questions Archive (Record #74544)