Thermal Properties of Matter
- 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 spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be :
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Given r1 = 12 cm , r2 = 6 cm
T1 = 500 K and T2 = 2 × 500 = 1000 K
P1 = 450 watt
Rate of power loss P ∝ r² T4P1 = r1²T14 P2 r2²T24 P2 = P1 r2²T24 t1 - t14
Solving we get, P2 = 1800 wattCorrect Option: C
Given r1 = 12 cm , r2 = 6 cm
T1 = 500 K and T2 = 2 × 500 = 1000 K
P1 = 450 watt
Rate of power loss P ∝ r² T4P1 = r1²T14 P2 r2²T24 P2 = P1 r2²T24 t1 - t14
Solving we get, P2 = 1800 watt
- A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U1, at wavelength 500 nm is U22 and that at 1000 nm is U3. Wien's constant, b = 2.88 × 106 nmK. Which of the following is correct ?
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According to wein's displacement law, maximum amount of emitted radiation corresponding to λm = b/T
λm = 2.88 × 106 nmk = 500nm 5760 k 
From the graph U1 < U2 < U3Correct Option: D
According to wein's displacement law, maximum amount of emitted radiation corresponding to λm = b/T
λm = 2.88 × 106 nmk = 500nm 5760 k
From the graph U1 < U2 < U3
- 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.
- Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K1 and K2. The thermal conductivity of the composite rod will be
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Heat current H = H1 + H2
= K1A(T1 - T2) + K2A(T1 - T2) d d = KEQ2A(T1 - T2) = A(T1 - T2) [k1 + k2] d d
Hence equivalent thermal conductivities for two rods of equal area is given byKEQ = k1 + k2 2 Correct Option: D
Heat current H = H1 + H2
= K1A(T1 - T2) + K2A(T1 - T2) d d = KEQ2A(T1 - T2) = A(T1 - T2) [k1 + k2] d d
Hence equivalent thermal conductivities for two rods of equal area is given byKEQ = k1 + k2 2
