Verified Solution • 2025–26 Session
very hard
QJEEA-2024-08: JEE Advanced 2024 JEEA-2024-08 Solution: Q.8 The specific heat capacity of a substance is temperature…
Q.8 The specific heat capacity of a substance is temperature dependent and is given by the formula
𝐶 = 𝑘𝑇, where 𝑘 is a constant of suitable dimensions in SI units, and 𝑇 is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from −73 °C to 27 °C is 𝑛𝑘, the value of 𝑛 is .
[Given: 0 K = −273 °C.]
Step-by-Step Verified Solution
Solution: 2500
- Convert Temperatures to Kelvin:
- T₁ = -73 °C = -73 + 273 K = 200 K
- T₂ = 27 °C = 27 + 273 K = 300 K
- Heat Required (dQ):
- dQ = mC dT, where m is the mass, C is the specific heat capacity, and dT is the change in temperature.
- Given C = kT, we have dQ = m(kT) dT.
- Integrate to Find Total Heat (Q):
- Q = ∫dQ = ∫m(kT) dT
- Q = m k ∫T dT (from T₁ to T₂)
- Q = mk [T²/2] (from 200 K to 300 K)
- Q = mk [(300² / 2) - (200² / 2)]
- Q = mk [(90000 / 2) - (40000 / 2)]
- Q = mk [(90000 - 40000) / 2]
- Q = mk [50000 / 2]
- Q = mk (25000)
- Given Mass (m):
- m = 1 kg
- Q = 1 * k * 25000 = 25000k
- Compare with nk:
- Q = nk
- 25000k = nk
- n = 25000
Answer:
The value of n is 25000.