Thermal Properties of Matter Moderate Questions and Answers | Page - 4

Thermal Properties of Matter

Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time?

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Initially liquid oxygen will gain the temperature up to its boiling temperature then it change its state to gas. After this again its temperature will increase, so corresponding graph will be

Correct Option: A

Initially liquid oxygen will gain the temperature up to its boiling temperature then it change its state to gas. After this again its temperature will increase, so corresponding graph will be

A slab of stone of area 0.36 m² and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is :
(Given latent heat of fusion of ice = 3.36 × 10⁵ J kg^–1 .) :

1. 1.24 J/m/s/°C
1. 1.29 J/m/s/°C
1. 2.05 J/m/s/°C
1. 1.02 J/m/s/°C

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Rate of heat given by steam = Rate of heat taken by ice
where K = Thermal conductivity of the slab
m = Mass of the ice
L = Latent heat of melting/fusion
A = Area of the slab
DQ = KA(100 - 0) = m dL ,
dt 1 dt

K × 100 × 0.36 = 4.8 × 3.36 × 10⁵
0.1 60 × 60

K = 1.24 j/m/s/°C

Correct Option: A

Rate of heat given by steam = Rate of heat taken by ice
where K = Thermal conductivity of the slab
m = Mass of the ice
L = Latent heat of melting/fusion
A = Area of the slab
DQ = KA(100 - 0) = m dL ,
dt 1 dt

K × 100 × 0.36 = 4.8 × 3.36 × 10⁵
0.1 60 × 60

K = 1.24 j/m/s/°C

A cylindrical metallic rod in therrnal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?

1. Q/4
1. Q/16
1. 2Q
1. Q/2

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The rate of heat flow is given by
Q = K.A ΔT
t l

Area of Original rod A = πR²;
Areal of new rod A′ = πR²/4.
Volume of original rod will be equal to the volume of new rod.
∴ πR²l = π R ² l'
2

⇒ l'/l = R² = 4
R²
4

∴ Q' = A' l = 1 , 1 = 1
Q A l' 4 4 16

∴ Q' = Q
16

Correct Option: B

The rate of heat flow is given by
Q = K.A ΔT
t l

Area of Original rod A = πR²;
Areal of new rod A′ = πR²/4.
Volume of original rod will be equal to the volume of new rod.
∴ πR²l = π R ² l'
2

⇒ l'/l = R² = 4
R²
4

∴ Q' = A' l = 1 , 1 = 1
Q A l' 4 4 16

∴ Q' = Q
16

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂). The rate of heat transfer, through the rod in a steady state is given by:

1. dQ = K(T₁ - T₂)
  dt LA
1. dQ = kLA (T₁ - T₂)
  dt
1. dQ = KA(T₁ - T₂)
  dt L
1. dQ = KL(T₁ - T₂)
  dt A

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dQ = kA(T₁ - T₂)
dt L

[(T₁ - T₂) is the temperature difference]

Correct Option: C

dQ = kA(T₁ - T₂)
dt L

[(T₁ - T₂) is the temperature difference]

A black body at 227°C radiates heat at the rate of 7 cals/cm²s. At a temperature of 727°C, the rate of heat radiated in the same units will be:

1. 50
1. 112
1. 80
1. 60

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According to Stefan’s law E = σT⁴,
T₁ = 500 K
T₂ = 1000 K

E₂ = T₂ ⁴ = 1000 ⁴ = 16
E₁ T₁ 500

∴ E₂ = 16 × 7 = 112 cal / cm²s

Correct Option: B

According to Stefan’s law E = σT⁴,
T₁ = 500 K
T₂ = 1000 K

E₂ = T₂ ⁴ = 1000 ⁴ = 16
E₁ T₁ 500

∴ E₂ = 16 × 7 = 112 cal / cm²s