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An aluminium container of mass 100 g holds 200 g of ice at -20 deg C. Heat is supplied to the whole system at a steady rate of 100 cal/s. Find the temperature of the system after 4 minutes. (specific heat of ice = 0.5 cal/g/deg C, latent heat of fusion of ice L = 80 cal/g, specific heat of aluminium = 0.2 cal/g/deg C, specific heat of water = 1 cal/g/deg C)
- About 25 deg C
- 0 deg C
- About 50 deg C
- 100 deg C
Correct answer: About 25 deg C
Solution
In 4 minutes the system receives 100 * 240 = 24000 cal. Warming the ice and aluminium from -20 to 0 deg C and then melting all the ice uses 2400 + 16000 = 18400 cal. The leftover 5600 cal warms the 200 g of water plus the 100 g aluminium, raising the temperature by about 25 deg C above 0 deg C.
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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