What is the mass of a material, whose specific heat capacity is 400 J/(kg °C) for a rise in temperature from 15 °C to 25 °C, when heat received is 20 kJ?
[amp_mcq option1=”0.1 kg” option2=”1 kg” option3=”10 kg” option4=”5 kg” correct=”option4″]
This question was previously asked in
UPSC NDA-1 – 2022
The formula used is Q = mcΔT, where Q is the heat energy absorbed (20 kJ = 20,000 J), m is the mass, c is the specific heat capacity (400 J/(kg °C)), and ΔT is the change in temperature (25 °C – 15 °C = 10 °C). Rearranging the formula to find mass: m = Q / (cΔT). Plugging in the values: m = 20000 J / (400 J/(kg °C) * 10 °C) = 20000 / 4000 kg = 5 kg.