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PHXI11:THERMAL PROPERTIES OF MATTER

366359 If mass energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

1 Remain unchanged

2 Increase

3 First increase than decrease

4 Decrease

Explanation:

When water is cooled at $0^{\circ} C$ to form ice then $80 c a l o r i e / g m$ (latent heat) energy is released. Because potential energy of the molecules decreases. Mass will decrease in the process of freezing of water.

PHXI11:THERMAL PROPERTIES OF MATTER

366360 A child running at a temperature of $101^{\circ} F$ is given an antipyrin (i.e., medicine that lower fever)which causes an increase in the rate of evaporation of sweet from his body. If the fever is brought down to $97^{\circ} F$ in $20 min$ , what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is $30 k g$ . The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about $580 c a l / g$ .

1

4.31 g / min

2

4.31 g / s

3

2.31 g / min

4

2.31 g / s

Explanation:

Given, mass of the child $(m) = 30 k g$
Time taken $(t) = 20 min$
Fall in temperature $= (101 - 98)^{\circ} F$
$Δ T = 3^{\circ} F = 3 \times \frac{5}{9}^{\circ} C = \frac{5}{3}^{\circ} C$
Specific heat of human body
$(s) = 4.2 \times 10^{3} J / k g -^{\circ} C$
Latent heat of evaporation
$(L) = 580 c a l / g = (580 \times 10^{3} \times 4.2) J / k g$
Heat given by body during fall in temperature
$Q_{1} = m s Δ T$
Let $m^{'}$ be the mass of sweat evaporates from the human body.
Heat required for evaporation
$Q_{2} = m^{'} L$
But
$Q_{1} = Q_{2} \Rightarrow m s Δ T = m^{'} L$
$m^{'} = \frac{m s Δ T}{L} = \frac{30 \times 4.2 \times 10^{3} \times 5 / 3}{580 \times 4.2 \times 10^{3}}$
$= \frac{10}{116} = 0.0862 k g$
Rate of evaporation of sweat $= \frac{0.0862}{20}$
$= 0.00431 k g / min$
$= 4.31 g / min$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366361 If $10 g$ of water is evaporated from $1 m^{2}$ field in one second then heat radiation incident is

1

280 \frac{c a l}{s}

2

5400 \frac{c a l}{s}

3

54 \frac{c a l}{s}

4

540 \frac{c a l}{s}

Explanation:

The total heat absorbed by water.
$H = m L$
The radiation power is $P = \frac{10}{1} \times 540 = 5400 \frac{c a l}{s}$

PHXI11:THERMAL PROPERTIES OF MATTER

366362 To raise the temperature of $200 g$ of ice at $0^{\circ} C$ to $10^{\circ} C$ by a heater of $420 W$ , the time required is $(1 C a l = 4.2 J)$

1

90 m i n

2

180 s e c o n d s

3

21.2 \min

4

21 .2 s e c o n d s

Explanation:

Total heat required is $Q = m L + m c Δ T$
$= 200 \times 80 + 200 \times 1 \times 10$
$Q = 18000 c a l = 75, 600 J$
Time required is
$Δ t = \frac{Q}{p} = \frac{75, 600}{420} = 180 s$

PHXI11:THERMAL PROPERTIES OF MATTER

366359 If mass energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

1 Remain unchanged

2 Increase

3 First increase than decrease

4 Decrease

Explanation:

When water is cooled at $0^{\circ} C$ to form ice then $80 c a l o r i e / g m$ (latent heat) energy is released. Because potential energy of the molecules decreases. Mass will decrease in the process of freezing of water.

PHXI11:THERMAL PROPERTIES OF MATTER

366360 A child running at a temperature of $101^{\circ} F$ is given an antipyrin (i.e., medicine that lower fever)which causes an increase in the rate of evaporation of sweet from his body. If the fever is brought down to $97^{\circ} F$ in $20 min$ , what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is $30 k g$ . The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about $580 c a l / g$ .

1

4.31 g / min

2

4.31 g / s

3

2.31 g / min

4

2.31 g / s

Explanation:

Given, mass of the child $(m) = 30 k g$
Time taken $(t) = 20 min$
Fall in temperature $= (101 - 98)^{\circ} F$
$Δ T = 3^{\circ} F = 3 \times \frac{5}{9}^{\circ} C = \frac{5}{3}^{\circ} C$
Specific heat of human body
$(s) = 4.2 \times 10^{3} J / k g -^{\circ} C$
Latent heat of evaporation
$(L) = 580 c a l / g = (580 \times 10^{3} \times 4.2) J / k g$
Heat given by body during fall in temperature
$Q_{1} = m s Δ T$
Let $m^{'}$ be the mass of sweat evaporates from the human body.
Heat required for evaporation
$Q_{2} = m^{'} L$
But
$Q_{1} = Q_{2} \Rightarrow m s Δ T = m^{'} L$
$m^{'} = \frac{m s Δ T}{L} = \frac{30 \times 4.2 \times 10^{3} \times 5 / 3}{580 \times 4.2 \times 10^{3}}$
$= \frac{10}{116} = 0.0862 k g$
Rate of evaporation of sweat $= \frac{0.0862}{20}$
$= 0.00431 k g / min$
$= 4.31 g / min$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366361 If $10 g$ of water is evaporated from $1 m^{2}$ field in one second then heat radiation incident is

1

280 \frac{c a l}{s}

2

5400 \frac{c a l}{s}

3

54 \frac{c a l}{s}

4

540 \frac{c a l}{s}

Explanation:

The total heat absorbed by water.
$H = m L$
The radiation power is $P = \frac{10}{1} \times 540 = 5400 \frac{c a l}{s}$

PHXI11:THERMAL PROPERTIES OF MATTER

366362 To raise the temperature of $200 g$ of ice at $0^{\circ} C$ to $10^{\circ} C$ by a heater of $420 W$ , the time required is $(1 C a l = 4.2 J)$

1

90 m i n

2

180 s e c o n d s

3

21.2 \min

4

21 .2 s e c o n d s

Explanation:

Total heat required is $Q = m L + m c Δ T$
$= 200 \times 80 + 200 \times 1 \times 10$
$Q = 18000 c a l = 75, 600 J$
Time required is
$Δ t = \frac{Q}{p} = \frac{75, 600}{420} = 180 s$

PHXI11:THERMAL PROPERTIES OF MATTER

366359 If mass energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

1 Remain unchanged

2 Increase

3 First increase than decrease

4 Decrease

Explanation:

When water is cooled at $0^{\circ} C$ to form ice then $80 c a l o r i e / g m$ (latent heat) energy is released. Because potential energy of the molecules decreases. Mass will decrease in the process of freezing of water.

PHXI11:THERMAL PROPERTIES OF MATTER

366360 A child running at a temperature of $101^{\circ} F$ is given an antipyrin (i.e., medicine that lower fever)which causes an increase in the rate of evaporation of sweet from his body. If the fever is brought down to $97^{\circ} F$ in $20 min$ , what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is $30 k g$ . The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about $580 c a l / g$ .

1

4.31 g / min

2

4.31 g / s

3

2.31 g / min

4

2.31 g / s

Explanation:

Given, mass of the child $(m) = 30 k g$
Time taken $(t) = 20 min$
Fall in temperature $= (101 - 98)^{\circ} F$
$Δ T = 3^{\circ} F = 3 \times \frac{5}{9}^{\circ} C = \frac{5}{3}^{\circ} C$
Specific heat of human body
$(s) = 4.2 \times 10^{3} J / k g -^{\circ} C$
Latent heat of evaporation
$(L) = 580 c a l / g = (580 \times 10^{3} \times 4.2) J / k g$
Heat given by body during fall in temperature
$Q_{1} = m s Δ T$
Let $m^{'}$ be the mass of sweat evaporates from the human body.
Heat required for evaporation
$Q_{2} = m^{'} L$
But
$Q_{1} = Q_{2} \Rightarrow m s Δ T = m^{'} L$
$m^{'} = \frac{m s Δ T}{L} = \frac{30 \times 4.2 \times 10^{3} \times 5 / 3}{580 \times 4.2 \times 10^{3}}$
$= \frac{10}{116} = 0.0862 k g$
Rate of evaporation of sweat $= \frac{0.0862}{20}$
$= 0.00431 k g / min$
$= 4.31 g / min$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366361 If $10 g$ of water is evaporated from $1 m^{2}$ field in one second then heat radiation incident is

1

280 \frac{c a l}{s}

2

5400 \frac{c a l}{s}

3

54 \frac{c a l}{s}

4

540 \frac{c a l}{s}

Explanation:

The total heat absorbed by water.
$H = m L$
The radiation power is $P = \frac{10}{1} \times 540 = 5400 \frac{c a l}{s}$

PHXI11:THERMAL PROPERTIES OF MATTER

366362 To raise the temperature of $200 g$ of ice at $0^{\circ} C$ to $10^{\circ} C$ by a heater of $420 W$ , the time required is $(1 C a l = 4.2 J)$

1

90 m i n

2

180 s e c o n d s

3

21.2 \min

4

21 .2 s e c o n d s

Explanation:

Total heat required is $Q = m L + m c Δ T$
$= 200 \times 80 + 200 \times 1 \times 10$
$Q = 18000 c a l = 75, 600 J$
Time required is
$Δ t = \frac{Q}{p} = \frac{75, 600}{420} = 180 s$

PHXI11:THERMAL PROPERTIES OF MATTER

366359 If mass energy equivalence is taken into account, when water is cooled to form ice, the mass of water should

1 Remain unchanged

2 Increase

3 First increase than decrease

4 Decrease

Explanation:

When water is cooled at $0^{\circ} C$ to form ice then $80 c a l o r i e / g m$ (latent heat) energy is released. Because potential energy of the molecules decreases. Mass will decrease in the process of freezing of water.

PHXI11:THERMAL PROPERTIES OF MATTER

366360 A child running at a temperature of $101^{\circ} F$ is given an antipyrin (i.e., medicine that lower fever)which causes an increase in the rate of evaporation of sweet from his body. If the fever is brought down to $97^{\circ} F$ in $20 min$ , what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is $30 k g$ . The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about $580 c a l / g$ .

1

4.31 g / min

2

4.31 g / s

3

2.31 g / min

4

2.31 g / s

Explanation:

Given, mass of the child $(m) = 30 k g$
Time taken $(t) = 20 min$
Fall in temperature $= (101 - 98)^{\circ} F$
$Δ T = 3^{\circ} F = 3 \times \frac{5}{9}^{\circ} C = \frac{5}{3}^{\circ} C$
Specific heat of human body
$(s) = 4.2 \times 10^{3} J / k g -^{\circ} C$
Latent heat of evaporation
$(L) = 580 c a l / g = (580 \times 10^{3} \times 4.2) J / k g$
Heat given by body during fall in temperature
$Q_{1} = m s Δ T$
Let $m^{'}$ be the mass of sweat evaporates from the human body.
Heat required for evaporation
$Q_{2} = m^{'} L$
But
$Q_{1} = Q_{2} \Rightarrow m s Δ T = m^{'} L$
$m^{'} = \frac{m s Δ T}{L} = \frac{30 \times 4.2 \times 10^{3} \times 5 / 3}{580 \times 4.2 \times 10^{3}}$
$= \frac{10}{116} = 0.0862 k g$
Rate of evaporation of sweat $= \frac{0.0862}{20}$
$= 0.00431 k g / min$
$= 4.31 g / min$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366361 If $10 g$ of water is evaporated from $1 m^{2}$ field in one second then heat radiation incident is

1

280 \frac{c a l}{s}

2

5400 \frac{c a l}{s}

3

54 \frac{c a l}{s}

4

540 \frac{c a l}{s}

Explanation:

The total heat absorbed by water.
$H = m L$
The radiation power is $P = \frac{10}{1} \times 540 = 5400 \frac{c a l}{s}$

PHXI11:THERMAL PROPERTIES OF MATTER

366362 To raise the temperature of $200 g$ of ice at $0^{\circ} C$ to $10^{\circ} C$ by a heater of $420 W$ , the time required is $(1 C a l = 4.2 J)$

1

90 m i n

2

180 s e c o n d s

3

21.2 \min

4

21 .2 s e c o n d s

Explanation:

Total heat required is $Q = m L + m c Δ T$
$= 200 \times 80 + 200 \times 1 \times 10$
$Q = 18000 c a l = 75, 600 J$
Time required is
$Δ t = \frac{Q}{p} = \frac{75, 600}{420} = 180 s$

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