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A hot liquid in a container of negligible heat capacity cools at a rate of 3 K/min just before it begins to solidify. While solidifying, its temperature stays constant for 30 min. Assuming the rate of heat loss stays constant, find the ratio of the specific heat capacity of the liquid to the specific latent heat of fusion (in K⁻¹).
- 1/30
- 1/60
- 1/90
- 1/120
Correct answer: 1/90
Solution
Let the constant rate of heat loss be P. In the liquid phase just before freezing, this power produces a 3 K/min temperature drop: P = m*s*3 (per minute). During solidification the temperature is constant, so all heat removed over 30 min is latent: m*L = P*30. Substituting P = 3*m*s gives m*L = 3*m*s*30 = 90*m*s, hence s/L = 1/90 K⁻¹.
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
Group-I:
(P) 2000 K
(Q) 3000 K
(R) 5000 K
(S) 10000 K
Group-II:
(1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV.
(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
(4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K.
(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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- A continuous-flow calorimeter is used in two separate experiments to find the specific heat of a liquid. In the first trial, supplying 60 W raises the liquid temperature by 10 K. In the second trial, the power is doubled to 120 W, but the same 10 K rise is maintained by tripling the flow rate. Assuming heat loss to surroundings is the same in both trials, what is the power lost to the surroundings?
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