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.