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Physics Thermal Properties of Matter

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Physics Thermal Properties of Matter Class 11th

For an ideal gas PT³ = constant. Find the coefficient of volume expansion of the gas is:

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Vishal Baghel

Contributor-Level 10

PT³ = k (constant)

$\Rightarrow (\frac{n R T}{V}) T^{3} = k$

$\Rightarrow T^{4} V^{- 1} = k$

$\Rightarrow T^{4} = k V$

$\Rightarrow 4 * \frac{Δ T}{T} = \frac{Δ V}{V}$ - (i) >

$\Rightarrow Δ V = V Y Δ T$ - (ii) >

Comparing I > & ii >

$\Rightarrow Y = \frac{4}{T}$

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New answer posted

a month ago

Physics Thermal Properties of Matter Class 11th

A balloon carries a total load of 185 kg of normal pressure and temperature of 27°C. What load with the balloon carry on rising to a height at which the barometer pressure is 45cm of Hg and the temperature.

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Vishal Baghel

Contributor-Level 10

$P = ρ R T$

$\Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{P_{1}}{P_{2}} * \frac{T_{2}}{T_{1}} = \frac{76}{45} * \frac{260}{300}$

$\Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{M_{1}}{M_{2}}$

$\Rightarrow M_{2} = \frac{45 * 300 * 185}{76 * 266} = 123.54 k g$

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

The temperature of equals masses of three different liquids x, y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. Temperature of mixture when x and z are mixed will be:

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alok kumar singh

Contributor-Level 10

ms_x (16 – 10) = ms_y (20 – 16)

$\Rightarrow \frac{s_{x}}{s_{y}} = \frac{2}{3}$

$m s_{y} (26 - 20) = m s_{z} (30 - 26)$

$\Rightarrow \frac{s_{y}}{s_{z}} = \frac{2}{3}$

$\frac{s_{x}}{s_{z}} = \frac{s_{x}}{s_{y}} * \frac{s_{y}}{s_{z}} = \frac{2}{3} * \frac{2}{3} = \frac{4}{9}$

$m s_{x} (T - 10) = m s_{z} (30 - T)$

->4(T – 10) = 9(30 – T)

->T = 23.84°C

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

Read the following statements:

(A) When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.

(B) Two bodies P and Q having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by P and Q are in the ratio 1:1. 15.

(C) A Carnot Engine working between 100K and 400K has an efficiency of 75%.

...more

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alok kumar singh

Contributor-Level 10

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{100}{400} = \frac{3}{4} = 75 %$

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

If K₁ and K₂ are the thermal conductivities, L₁ and L₂ are the lengths and A₁ and A₂ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2 .$

Then, for the arrangement as shown in the figure, the value of temperature T of the steel – copper junction in the steady state will be:

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alok kumar singh

Contributor-Level 10

$\frac{K_{2}}{k_{1}} = 9 \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$

$\frac{450 - T}{\frac{L_{1}}{A_{1} k_{1}}} = \frac{T}{\frac{L_{2}}{k_{2} A_{2}}} \Rightarrow (450 - T) \frac{L_{2}}{K_{2} A_{2}} = (T) \frac{L_{1}}{A_{1} k_{1}}$

$(450 - T) = T (\frac{L_{1}}{L_{2}} \frac{k_{2} A_{2}}{k_{1} A_{1}})$

$\Rightarrow 450 - T = T (2 * 9 * \frac{1}{2}) = 9 T$

10 T = 450

T = 45° C

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

When a gas filled in a closed is heated by raising the temperature by $1 ° C$ , its pressure increases by 0.4%. The initial temperature of the gas is……….K.

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alok kumar singh

Contributor-Level 10

For closed vessel

P a T [v = constant]

->P = kT

$\frac{0.4}{100} = \frac{1}{1}$

T = 250 K

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

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alok kumar singh

Contributor-Level 10

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

->5 * 0.39 * 500 = m_ice * 335

$m_{i c e} = \frac{5 * 0.39 * 500}{335}$

= 2.91 kg

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

A solid metallic cube having total surface area 24 m² is uniformly heated. if its temperature is increased by 10⁰C, calculate the increase in volume of the cube.

(Given a = 5.0 * 10^-4 ⁰C^-1).

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alok kumar singh

Contributor-Level 10

Given

6a² = 24

a² = 4

a = 2m

$Δ v = v (γ) Δ T$

$= 3 α v Δ T$

$= 3 * 5 * 1 0^{- 4} * (a^{3}) * 10$

$= 3 * 5 * 1 0^{- 4} * 8 * 10$

$\begin{array}{l} = 120 * 1 0^{3} \\ = 1.2 * 1 0^{5} c m^{3} \end{array}$

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

Read the following statements:

(A) When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.

(B) Two bodies P and Q having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by P and Q are in the ratio 1:1. 15.

(C) A Carnot Engine working between 100K and 400K has an efficiency of 75%.

...more

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

A.A is correct statement

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{100}{400} = \frac{3}{4} = 75 %$

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New answer posted

2 months ago

Physics Thermal Properties of Matter Class 11th

If K₁ and K₂ are the thermal conductivities, L₁ and L₂ are the lengths and A₁ and A₂ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2 .$ Then, for the arrangement as shown in the figure, the value of temperature T of the steel – copper junction in the steady state will be:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$\frac{K_{2}}{k_{1}} = 9 \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$

$\frac{450 - T}{\frac{L_{1}}{A_{1} k_{1}}} = \frac{T}{\frac{L_{2}}{k_{2} A_{2}}} \Rightarrow (450 - T) \frac{L_{2}}{K_{2} A_{2}} = (T) \frac{L_{1}}{A_{1} k_{1}}$

$(450 - T) = T (\frac{L_{1}}{L_{2}} \frac{k_{2} A_{2}}{k_{1} A_{1}})$ $\Rightarrow 450 - T = T (2 * 9 * \frac{1}{2}) = 9 T$

10 T = 450

T = 45° C

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