Q: 1 (NDA-I/2022)

SET A . Unofficial Key

question_subject:

Science

question_exam:

NDA-I

stats:

0,30,67,15,25,27,30

keywords:

{'specific heat capacity': [0, 0, 0, 2], 'temperature': [0, 1, 1, 7], 'mass': [0, 0, 2, 3], 'kg': [0, 1, 9, 24], 'heat': [10, 3, 13, 46], 'material': [0, 0, 0, 1], 'kj': [0, 1, 1, 1]}

We can use the formula:

Q = mc?T

where Q is the heat received, m is the mass of the material, c is the specific heat capacity, and ?T is the change in temperature.

We know that Q = 20 kJ = 20,000 J, c = 400 J/(kg°C), and ?T = 25 °C - 15 °C = 10 °C.

Substituting these values into the formula, we get:

20,000 J = m × 400 J/(kg°C) × 10 °C

Simplifying the equation, we can cancel out the units of °C, and solve for m:

m = 20,000 J / (400 J/(kg°C) × 10) = 5 kg

Therefore, the mass of the material is 5 kg.