Thermal Properties of Matter
- Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is:
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Let the temperature of surroundings be θ0
By Newton's law of coolingθ1 - θ2 = k θ1 + θ2 - θ0 t 2 ⇒ 70 - 60 = k 70 + 60 - θ0 5 2
⇒ 2 = k [65 – θ0] ...(i)Similarly, 60 - 54 = k 60 + 54 - θ0 5 2 ⇒ 6 = k[57 - θ0] .............(ii) 5
By dividing (i) by (ii) we have10 = 65 - θ0 ⇒ θ0 = 45° 6 57 - θ0 Correct Option: A
Let the temperature of surroundings be θ0
By Newton's law of coolingθ1 - θ2 = k θ1 + θ2 - θ0 t 2 ⇒ 70 - 60 = k 70 + 60 - θ0 5 2
⇒ 2 = k [65 – θ0] ...(i)Similarly, 60 - 54 = k 60 + 54 - θ0 5 2 ⇒ 6 = k[57 - θ0] .............(ii) 5
By dividing (i) by (ii) we have10 = 65 - θ0 ⇒ θ0 = 45° 6 57 - θ0
- If λm denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then
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From Wein’s displacement law
λmT = constant
⇒ λm ∝ T–1Correct Option: A
From Wein’s displacement law
λmT = constant
⇒ λm ∝ T–1
- Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is
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In series, equivalent thermal conductivity
Keq = 2K1K2 K1 + K2 or, Keq = 2 × K × 2K = 4 K K + 2K 3 Correct Option: A
In series, equivalent thermal conductivity
Keq = 2K1K2 K1 + K2 or, Keq = 2 × K × 2K = 4 K K + 2K 3
- Wien's law is concerned with
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According to Wein's displacement law, product of wavelength belonging to maximum intensity and temperature is constant i.e., λmT = constant.
Correct Option: D
According to Wein's displacement law, product of wavelength belonging to maximum intensity and temperature is constant i.e., λmT = constant.
- Two rods of thermal conductivities K1 and K2, cross-sections A1 and A2 and specific heats S1 and S2 are of equal lengths. The temperatures of two ends of each rod are T1 and T2. The rate of flow of heat at the steady state will be equal if
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Rate of heat flow for one rod = K1A1(T1 - T2) (d → Length) d
Rate of heat flow for other rod= K2A2(T1 - T2) d In steady state, K1A1(T1 - T2) d = K2A2(T1 - T2) ⇒ K1A1 = K2A2 d Correct Option: B
Rate of heat flow for one rod = K1A1(T1 - T2) (d → Length) d
Rate of heat flow for other rod= K2A2(T1 - T2) d In steady state, K1A1(T1 - T2) d = K2A2(T1 - T2) ⇒ K1A1 = K2A2 d