Thermal Properties of Matter
- The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be
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As the temperature difference ∆T = 10°C as well as the thermal resistance is same for both the cases, so thermal current or rate of heat flow will also be same for both the cases.
Correct Option: C
As the temperature difference ∆T = 10°C as well as the thermal resistance is same for both the cases, so thermal current or rate of heat flow will also be same for both the cases.
- Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be:
[Take specific heat of water = 1 cal g–1 °C–1 and latent heat of steam = 540 cal g–1]
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According to the principle of calorimetry.
Heat lost = Heat gained
mLv + msw∆θ = mwswΔ θ
⇒ m × 540 + m × 1 × (100 – 80)
= 20 × 1 × (80 – 10)
⇒ m = 2.5 g
Therefore total mass of water at 80°C
= (20 + 2.5) g = 22.5 gCorrect Option: D
According to the principle of calorimetry.
Heat lost = Heat gained
mLv + msw∆θ = mwswΔ θ
⇒ m × 540 + m × 1 × (100 – 80)
= 20 × 1 × (80 – 10)
⇒ m = 2.5 g
Therefore total mass of water at 80°C
= (20 + 2.5) g = 22.5 g
- A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using
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Wein’s displacement law
According to this law
λmax ∝ 1/T
or, λmax × T = constant
So, as the temperature increases λ decreases.Correct Option: A
Wein’s displacement law
According to this law
λmax ∝ 1/T
or, λmax × T = constant
So, as the temperature increases λ decreases.
- Two metal rods 1 and 2 of same lengths have same temperature difference between their ends. Their thermal conductivities are K1 and K2 and cross sectional areas A1 and A2, respectively. If the rate of heat conduction in rod 1 is four times that in rod 2, then
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Q1 = 4Q2 (Given)
⇒ K1A1Δt = 4 K2A2Δt ⇒ K1A1 = 4K2A2 L L
Q = σeAT4Correct Option: B
Q1 = 4Q2 (Given)
⇒ K1A1Δt = 4 K2A2Δt ⇒ K1A1 = 4K2A2 L L
Q = σeAT4
- A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C? Assume the temperature of surroundings to be 30.0°C and Newton's law of cooling to be valid.
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Rate of cooling ∝ temperature difference
between system and surrounding.
As the temperature difference is halved, so the rate of cooling will also be halved.
So time taken will be doubled
t = 2 × 5 sec. = 10 sec.Correct Option: B
Rate of cooling ∝ temperature difference
between system and surrounding.
As the temperature difference is halved, so the rate of cooling will also be halved.
So time taken will be doubled
t = 2 × 5 sec. = 10 sec.