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PHXI11:THERMAL PROPERTIES OF MATTER

366337 Compared to a burn due to water at $100^{\circ} C$ , a burn due to steam at $100^{\circ} C$ is:

1 More dangerous

2 Less dangerous

3 Equally dangerous

4 None of these

Explanation:

Steam at $100^{\circ} C$ contains extra $2260 J / g m$ energy as compared to water at $100^{\circ} C$ . So it's more dangerous to burn with steam than water.

PHXI11:THERMAL PROPERTIES OF MATTER

366338 In an industrial process $10 k g$ of water per hour is to be heated from $20^{\circ} C$ to $80^{\circ} C$ . To do this steam at $150^{\circ} C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^{\circ} C$ . How many $k g$ of steam is required per hour? (Specific heat of steam $=$ specific heat of water $= 1 c a l / g^{\circ} C$ , Latent heat of vaporization $= 540 c a l / g$ )

1

1 k g

2

3 k g

3

6 k g

4

2 k g

Explanation:

Suppose $m k g$ steam is required per hour
Heat is released by steam in following three steps
(i) When $150^{\circ} C$ steam $\vec{Q_{1}} 100^{\circ} C$ steam $Q_{1} = m c_{steam} Δ θ = m \times 1 (150 - 100) = 50 m c a l$
(ii) When $100^{\circ} C$ steam $\vec{Q_{2}} 100^{\circ} C$ water
$Q_{2} = m L_{v} = m \times 540 = 540 m c a l$
(iii) When $100^{\circ} C$ water $\vec{Q_{2}} 90^{\circ} C$ water
$Q_{3} = m c_{w} Δ θ = m \times 1 \times (100 - 90) = 10 m c a l$
Hence total heat given by the steam
$Q = Q_{1} + Q_{2} + Q_{3} = 600 m c a l (1)$
Heat taken by $10 k g$ water
$Q^{'} m c_{w} Δ θ = 10 \times 10^{3} \times 1 \times (80 - 20)$
$= 600 \times 10^{3} c a l$
Hence $Q = Q^{'} \Rightarrow 600 m = 600 \times 10^{3}$
$\Rightarrow m = 10^{3} g = 1 k g$

PHXI11:THERMAL PROPERTIES OF MATTER

366339 An iron rocket fragment initially at $- 100^{\circ} C$ enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is $0.11 k c a l / k g^{\circ} C$ its melting point is $1535^{\circ} C$ and the latent heat of fusion is $3 k c a l / k g$ . The minimum velocity with which the fragment must have entered the atmosphere is

1

0.45 k m / s

2

1.32 k m / s

3

2.32 k m / s

4

Z e r o

Explanation:

The entire kinetic energy of the fragment is changed to heat. Expressing mass $m$ in kilograms everywhere, we have
$\frac{1}{2} m v^{2} = [m (30) + m (0.11) (1535^{\circ} + 100^{\circ})] 4184$
$v^{2} = (8368) [30 + 180] ≃ 1.76 \times 10^{6}$
$\Rightarrow v = (\sqrt{1.76}) \times 10^{3} = 1.32 k m / s$

PHXI11:THERMAL PROPERTIES OF MATTER

366340 $2 g$ of steam condenses when passes through $40 g$ of water initially at $25^{\circ} C$ . The condensation of steam raises the temperature of water to ${54.3}^{\circ} C$ . What is the latent heat of steam

1

536 c a l / g

2

540 c a l / g

3

480 c a l / g

4

270 c a l / g

Explanation:

Let $L$ be the latent heat and using principle of calorimetry.
$2 L + 2 (100 - 54.3) = 40 \times (54.3 - 25.0)$
$\Rightarrow L = 540.3 c a l / g$

PHXI11:THERMAL PROPERTIES OF MATTER

366337 Compared to a burn due to water at $100^{\circ} C$ , a burn due to steam at $100^{\circ} C$ is:

1 More dangerous

2 Less dangerous

3 Equally dangerous

4 None of these

Explanation:

Steam at $100^{\circ} C$ contains extra $2260 J / g m$ energy as compared to water at $100^{\circ} C$ . So it's more dangerous to burn with steam than water.

PHXI11:THERMAL PROPERTIES OF MATTER

366338 In an industrial process $10 k g$ of water per hour is to be heated from $20^{\circ} C$ to $80^{\circ} C$ . To do this steam at $150^{\circ} C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^{\circ} C$ . How many $k g$ of steam is required per hour? (Specific heat of steam $=$ specific heat of water $= 1 c a l / g^{\circ} C$ , Latent heat of vaporization $= 540 c a l / g$ )

1

1 k g

2

3 k g

3

6 k g

4

2 k g

Explanation:

Suppose $m k g$ steam is required per hour
Heat is released by steam in following three steps
(i) When $150^{\circ} C$ steam $\vec{Q_{1}} 100^{\circ} C$ steam $Q_{1} = m c_{steam} Δ θ = m \times 1 (150 - 100) = 50 m c a l$
(ii) When $100^{\circ} C$ steam $\vec{Q_{2}} 100^{\circ} C$ water
$Q_{2} = m L_{v} = m \times 540 = 540 m c a l$
(iii) When $100^{\circ} C$ water $\vec{Q_{2}} 90^{\circ} C$ water
$Q_{3} = m c_{w} Δ θ = m \times 1 \times (100 - 90) = 10 m c a l$
Hence total heat given by the steam
$Q = Q_{1} + Q_{2} + Q_{3} = 600 m c a l (1)$
Heat taken by $10 k g$ water
$Q^{'} m c_{w} Δ θ = 10 \times 10^{3} \times 1 \times (80 - 20)$
$= 600 \times 10^{3} c a l$
Hence $Q = Q^{'} \Rightarrow 600 m = 600 \times 10^{3}$
$\Rightarrow m = 10^{3} g = 1 k g$

PHXI11:THERMAL PROPERTIES OF MATTER

366339 An iron rocket fragment initially at $- 100^{\circ} C$ enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is $0.11 k c a l / k g^{\circ} C$ its melting point is $1535^{\circ} C$ and the latent heat of fusion is $3 k c a l / k g$ . The minimum velocity with which the fragment must have entered the atmosphere is

1

0.45 k m / s

2

1.32 k m / s

3

2.32 k m / s

4

Z e r o

Explanation:

The entire kinetic energy of the fragment is changed to heat. Expressing mass $m$ in kilograms everywhere, we have
$\frac{1}{2} m v^{2} = [m (30) + m (0.11) (1535^{\circ} + 100^{\circ})] 4184$
$v^{2} = (8368) [30 + 180] ≃ 1.76 \times 10^{6}$
$\Rightarrow v = (\sqrt{1.76}) \times 10^{3} = 1.32 k m / s$

PHXI11:THERMAL PROPERTIES OF MATTER

366340 $2 g$ of steam condenses when passes through $40 g$ of water initially at $25^{\circ} C$ . The condensation of steam raises the temperature of water to ${54.3}^{\circ} C$ . What is the latent heat of steam

1

536 c a l / g

2

540 c a l / g

3

480 c a l / g

4

270 c a l / g

Explanation:

Let $L$ be the latent heat and using principle of calorimetry.
$2 L + 2 (100 - 54.3) = 40 \times (54.3 - 25.0)$
$\Rightarrow L = 540.3 c a l / g$

PHXI11:THERMAL PROPERTIES OF MATTER

366337 Compared to a burn due to water at $100^{\circ} C$ , a burn due to steam at $100^{\circ} C$ is:

1 More dangerous

2 Less dangerous

3 Equally dangerous

4 None of these

Explanation:

Steam at $100^{\circ} C$ contains extra $2260 J / g m$ energy as compared to water at $100^{\circ} C$ . So it's more dangerous to burn with steam than water.

PHXI11:THERMAL PROPERTIES OF MATTER

366338 In an industrial process $10 k g$ of water per hour is to be heated from $20^{\circ} C$ to $80^{\circ} C$ . To do this steam at $150^{\circ} C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^{\circ} C$ . How many $k g$ of steam is required per hour? (Specific heat of steam $=$ specific heat of water $= 1 c a l / g^{\circ} C$ , Latent heat of vaporization $= 540 c a l / g$ )

1

1 k g

2

3 k g

3

6 k g

4

2 k g

Explanation:

Suppose $m k g$ steam is required per hour
Heat is released by steam in following three steps
(i) When $150^{\circ} C$ steam $\vec{Q_{1}} 100^{\circ} C$ steam $Q_{1} = m c_{steam} Δ θ = m \times 1 (150 - 100) = 50 m c a l$
(ii) When $100^{\circ} C$ steam $\vec{Q_{2}} 100^{\circ} C$ water
$Q_{2} = m L_{v} = m \times 540 = 540 m c a l$
(iii) When $100^{\circ} C$ water $\vec{Q_{2}} 90^{\circ} C$ water
$Q_{3} = m c_{w} Δ θ = m \times 1 \times (100 - 90) = 10 m c a l$
Hence total heat given by the steam
$Q = Q_{1} + Q_{2} + Q_{3} = 600 m c a l (1)$
Heat taken by $10 k g$ water
$Q^{'} m c_{w} Δ θ = 10 \times 10^{3} \times 1 \times (80 - 20)$
$= 600 \times 10^{3} c a l$
Hence $Q = Q^{'} \Rightarrow 600 m = 600 \times 10^{3}$
$\Rightarrow m = 10^{3} g = 1 k g$

PHXI11:THERMAL PROPERTIES OF MATTER

366339 An iron rocket fragment initially at $- 100^{\circ} C$ enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is $0.11 k c a l / k g^{\circ} C$ its melting point is $1535^{\circ} C$ and the latent heat of fusion is $3 k c a l / k g$ . The minimum velocity with which the fragment must have entered the atmosphere is

1

0.45 k m / s

2

1.32 k m / s

3

2.32 k m / s

4

Z e r o

Explanation:

The entire kinetic energy of the fragment is changed to heat. Expressing mass $m$ in kilograms everywhere, we have
$\frac{1}{2} m v^{2} = [m (30) + m (0.11) (1535^{\circ} + 100^{\circ})] 4184$
$v^{2} = (8368) [30 + 180] ≃ 1.76 \times 10^{6}$
$\Rightarrow v = (\sqrt{1.76}) \times 10^{3} = 1.32 k m / s$

PHXI11:THERMAL PROPERTIES OF MATTER

366340 $2 g$ of steam condenses when passes through $40 g$ of water initially at $25^{\circ} C$ . The condensation of steam raises the temperature of water to ${54.3}^{\circ} C$ . What is the latent heat of steam

1

536 c a l / g

2

540 c a l / g

3

480 c a l / g

4

270 c a l / g

Explanation:

Let $L$ be the latent heat and using principle of calorimetry.
$2 L + 2 (100 - 54.3) = 40 \times (54.3 - 25.0)$
$\Rightarrow L = 540.3 c a l / g$

PHXI11:THERMAL PROPERTIES OF MATTER

366337 Compared to a burn due to water at $100^{\circ} C$ , a burn due to steam at $100^{\circ} C$ is:

1 More dangerous

2 Less dangerous

3 Equally dangerous

4 None of these

Explanation:

Steam at $100^{\circ} C$ contains extra $2260 J / g m$ energy as compared to water at $100^{\circ} C$ . So it's more dangerous to burn with steam than water.

PHXI11:THERMAL PROPERTIES OF MATTER

366338 In an industrial process $10 k g$ of water per hour is to be heated from $20^{\circ} C$ to $80^{\circ} C$ . To do this steam at $150^{\circ} C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^{\circ} C$ . How many $k g$ of steam is required per hour? (Specific heat of steam $=$ specific heat of water $= 1 c a l / g^{\circ} C$ , Latent heat of vaporization $= 540 c a l / g$ )

1

1 k g

2

3 k g

3

6 k g

4

2 k g

Explanation:

Suppose $m k g$ steam is required per hour
Heat is released by steam in following three steps
(i) When $150^{\circ} C$ steam $\vec{Q_{1}} 100^{\circ} C$ steam $Q_{1} = m c_{steam} Δ θ = m \times 1 (150 - 100) = 50 m c a l$
(ii) When $100^{\circ} C$ steam $\vec{Q_{2}} 100^{\circ} C$ water
$Q_{2} = m L_{v} = m \times 540 = 540 m c a l$
(iii) When $100^{\circ} C$ water $\vec{Q_{2}} 90^{\circ} C$ water
$Q_{3} = m c_{w} Δ θ = m \times 1 \times (100 - 90) = 10 m c a l$
Hence total heat given by the steam
$Q = Q_{1} + Q_{2} + Q_{3} = 600 m c a l (1)$
Heat taken by $10 k g$ water
$Q^{'} m c_{w} Δ θ = 10 \times 10^{3} \times 1 \times (80 - 20)$
$= 600 \times 10^{3} c a l$
Hence $Q = Q^{'} \Rightarrow 600 m = 600 \times 10^{3}$
$\Rightarrow m = 10^{3} g = 1 k g$

PHXI11:THERMAL PROPERTIES OF MATTER

366339 An iron rocket fragment initially at $- 100^{\circ} C$ enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is $0.11 k c a l / k g^{\circ} C$ its melting point is $1535^{\circ} C$ and the latent heat of fusion is $3 k c a l / k g$ . The minimum velocity with which the fragment must have entered the atmosphere is

1

0.45 k m / s

2

1.32 k m / s

3

2.32 k m / s

4

Z e r o

Explanation:

The entire kinetic energy of the fragment is changed to heat. Expressing mass $m$ in kilograms everywhere, we have
$\frac{1}{2} m v^{2} = [m (30) + m (0.11) (1535^{\circ} + 100^{\circ})] 4184$
$v^{2} = (8368) [30 + 180] ≃ 1.76 \times 10^{6}$
$\Rightarrow v = (\sqrt{1.76}) \times 10^{3} = 1.32 k m / s$

PHXI11:THERMAL PROPERTIES OF MATTER

366340 $2 g$ of steam condenses when passes through $40 g$ of water initially at $25^{\circ} C$ . The condensation of steam raises the temperature of water to ${54.3}^{\circ} C$ . What is the latent heat of steam

1

536 c a l / g

2

540 c a l / g

3

480 c a l / g

4

270 c a l / g

Explanation:

Let $L$ be the latent heat and using principle of calorimetry.
$2 L + 2 (100 - 54.3) = 40 \times (54.3 - 25.0)$
$\Rightarrow L = 540.3 c a l / g$

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