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PHXI11:THERMAL PROPERTIES OF MATTER

366286 Calculate the amount of heat (in calories) required to convert $5 g$ of ice at $0^{\circ} C$ to steam at $100^{\circ} C$

1

3200 c a l

2

3100 c a l

3

4200 c a l

4

3600 c a l

Explanation:

Ice $(0^{\circ} C)$ converts into steam $(100^{\circ} C)$ in three steps.
The heat required $Q = Q_{1} + Q_{2} + Q_{3}$
$= 5 \times 80 + 5 \times 1 \times (100 - 0) + 5 \times 540 = 3600 c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366287 In similar calorimeters, equal volumes of water and alcohol, when poured, take $100 \sec$ and 74 sec respectively, to cool from $50^{\circ} C$ to $40^{\circ} C$ . If the thermal capacity of each calorimeter is numerically equal to volume of either liquid, then calculate the specific heat capacity of alcohol (given: the relative density of alcohol as 0.8 and specific heat capacity of water as
1 cal g $^{- 10} C^{- 1})$ .

1

0.8 c a l / g m^{\circ} C

2

0.6 c a l / g m^{\circ} C

3

0.9 c a l / g m^{\circ} C

4

1 c a l / g m^{\circ} C

Explanation:

Let $V$ be volume of either liquid
Mass of water $= V \times 1 g m$
Mass of alcohol $= V \times 0.8 = 0.8 V g m$
Rate of cooling of the water calorimeter
$= \frac{1}{100} [V \times (50^{\circ} - 40^{\circ}) + V \times (50^{\circ} - 40^{\circ})]$
$= (1 / 5) V c a l / \sec$
Rate of cooling of alcohol calorimeter $= \frac{1}{74} [V \times [50^{\circ} - 40^{\circ}] + 0.8 V \times c \times (50^{\circ} - 40^{\circ})]]$
$= (1 / 74) (10 V + 8 V c) c a l / \sec$
It is given that calorimetrers are similar to as rate of cooling of both is same
$\frac{1}{5} V = (1 / 74) 10 V + 8 V c$
$c = 0.6 c a l / g m^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366288 A chef, on finding his stove out of order, decides to boil coffee by shaking it in a thermosflask. Suppose that he uses water at $15^{\circ} C$ and that the water falls $30 c m$ each shake, the chef making 30 shakes each minute. Neglecting any loss of thermal energy by the flask, how long must he shake the flask until the water reaches $100^{\circ} C (c_{c o f f e e} = 4190 J)$ ?

1

2.25 \times 10^{3} min

2

3.97 \times 10^{3} min

3

4.03 \times 10^{3} min

4

5.25 \times 10^{3} min

Explanation:

The time taken to boil the coffee is
Hence, $t = \frac{Q}{N m g h} = \frac{c m (T_{f} - T_{i})}{N m g h} = \frac{c (T_{f} - T_{i})}{N g h}$
$= \frac{(4190) (100 - 15)}{(30) (9.8) (0.30)} = 4.03 \times 10^{3} \min$

PHXI11:THERMAL PROPERTIES OF MATTER

366289 Which of the following states of matter have two specific heats?

1 Solid

2 Gas

3 Liquid

4 All three

Explanation:

For solids and liquids $C_{P} \approx C_{V}$ . Whereas for gases $C_{P} - C_{V} = R$

PHXI11:THERMAL PROPERTIES OF MATTER

366286 Calculate the amount of heat (in calories) required to convert $5 g$ of ice at $0^{\circ} C$ to steam at $100^{\circ} C$

1

3200 c a l

2

3100 c a l

3

4200 c a l

4

3600 c a l

Explanation:

Ice $(0^{\circ} C)$ converts into steam $(100^{\circ} C)$ in three steps.
The heat required $Q = Q_{1} + Q_{2} + Q_{3}$
$= 5 \times 80 + 5 \times 1 \times (100 - 0) + 5 \times 540 = 3600 c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366287 In similar calorimeters, equal volumes of water and alcohol, when poured, take $100 \sec$ and 74 sec respectively, to cool from $50^{\circ} C$ to $40^{\circ} C$ . If the thermal capacity of each calorimeter is numerically equal to volume of either liquid, then calculate the specific heat capacity of alcohol (given: the relative density of alcohol as 0.8 and specific heat capacity of water as
1 cal g $^{- 10} C^{- 1})$ .

1

0.8 c a l / g m^{\circ} C

2

0.6 c a l / g m^{\circ} C

3

0.9 c a l / g m^{\circ} C

4

1 c a l / g m^{\circ} C

Explanation:

Let $V$ be volume of either liquid
Mass of water $= V \times 1 g m$
Mass of alcohol $= V \times 0.8 = 0.8 V g m$
Rate of cooling of the water calorimeter
$= \frac{1}{100} [V \times (50^{\circ} - 40^{\circ}) + V \times (50^{\circ} - 40^{\circ})]$
$= (1 / 5) V c a l / \sec$
Rate of cooling of alcohol calorimeter $= \frac{1}{74} [V \times [50^{\circ} - 40^{\circ}] + 0.8 V \times c \times (50^{\circ} - 40^{\circ})]]$
$= (1 / 74) (10 V + 8 V c) c a l / \sec$
It is given that calorimetrers are similar to as rate of cooling of both is same
$\frac{1}{5} V = (1 / 74) 10 V + 8 V c$
$c = 0.6 c a l / g m^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366288 A chef, on finding his stove out of order, decides to boil coffee by shaking it in a thermosflask. Suppose that he uses water at $15^{\circ} C$ and that the water falls $30 c m$ each shake, the chef making 30 shakes each minute. Neglecting any loss of thermal energy by the flask, how long must he shake the flask until the water reaches $100^{\circ} C (c_{c o f f e e} = 4190 J)$ ?

1

2.25 \times 10^{3} min

2

3.97 \times 10^{3} min

3

4.03 \times 10^{3} min

4

5.25 \times 10^{3} min

Explanation:

The time taken to boil the coffee is
Hence, $t = \frac{Q}{N m g h} = \frac{c m (T_{f} - T_{i})}{N m g h} = \frac{c (T_{f} - T_{i})}{N g h}$
$= \frac{(4190) (100 - 15)}{(30) (9.8) (0.30)} = 4.03 \times 10^{3} \min$

PHXI11:THERMAL PROPERTIES OF MATTER

366289 Which of the following states of matter have two specific heats?

1 Solid

2 Gas

3 Liquid

4 All three

Explanation:

For solids and liquids $C_{P} \approx C_{V}$ . Whereas for gases $C_{P} - C_{V} = R$

PHXI11:THERMAL PROPERTIES OF MATTER

366286 Calculate the amount of heat (in calories) required to convert $5 g$ of ice at $0^{\circ} C$ to steam at $100^{\circ} C$

1

3200 c a l

2

3100 c a l

3

4200 c a l

4

3600 c a l

Explanation:

Ice $(0^{\circ} C)$ converts into steam $(100^{\circ} C)$ in three steps.
The heat required $Q = Q_{1} + Q_{2} + Q_{3}$
$= 5 \times 80 + 5 \times 1 \times (100 - 0) + 5 \times 540 = 3600 c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366287 In similar calorimeters, equal volumes of water and alcohol, when poured, take $100 \sec$ and 74 sec respectively, to cool from $50^{\circ} C$ to $40^{\circ} C$ . If the thermal capacity of each calorimeter is numerically equal to volume of either liquid, then calculate the specific heat capacity of alcohol (given: the relative density of alcohol as 0.8 and specific heat capacity of water as
1 cal g $^{- 10} C^{- 1})$ .

1

0.8 c a l / g m^{\circ} C

2

0.6 c a l / g m^{\circ} C

3

0.9 c a l / g m^{\circ} C

4

1 c a l / g m^{\circ} C

Explanation:

Let $V$ be volume of either liquid
Mass of water $= V \times 1 g m$
Mass of alcohol $= V \times 0.8 = 0.8 V g m$
Rate of cooling of the water calorimeter
$= \frac{1}{100} [V \times (50^{\circ} - 40^{\circ}) + V \times (50^{\circ} - 40^{\circ})]$
$= (1 / 5) V c a l / \sec$
Rate of cooling of alcohol calorimeter $= \frac{1}{74} [V \times [50^{\circ} - 40^{\circ}] + 0.8 V \times c \times (50^{\circ} - 40^{\circ})]]$
$= (1 / 74) (10 V + 8 V c) c a l / \sec$
It is given that calorimetrers are similar to as rate of cooling of both is same
$\frac{1}{5} V = (1 / 74) 10 V + 8 V c$
$c = 0.6 c a l / g m^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366288 A chef, on finding his stove out of order, decides to boil coffee by shaking it in a thermosflask. Suppose that he uses water at $15^{\circ} C$ and that the water falls $30 c m$ each shake, the chef making 30 shakes each minute. Neglecting any loss of thermal energy by the flask, how long must he shake the flask until the water reaches $100^{\circ} C (c_{c o f f e e} = 4190 J)$ ?

1

2.25 \times 10^{3} min

2

3.97 \times 10^{3} min

3

4.03 \times 10^{3} min

4

5.25 \times 10^{3} min

Explanation:

The time taken to boil the coffee is
Hence, $t = \frac{Q}{N m g h} = \frac{c m (T_{f} - T_{i})}{N m g h} = \frac{c (T_{f} - T_{i})}{N g h}$
$= \frac{(4190) (100 - 15)}{(30) (9.8) (0.30)} = 4.03 \times 10^{3} \min$

PHXI11:THERMAL PROPERTIES OF MATTER

366289 Which of the following states of matter have two specific heats?

1 Solid

2 Gas

3 Liquid

4 All three

Explanation:

For solids and liquids $C_{P} \approx C_{V}$ . Whereas for gases $C_{P} - C_{V} = R$

PHXI11:THERMAL PROPERTIES OF MATTER

366286 Calculate the amount of heat (in calories) required to convert $5 g$ of ice at $0^{\circ} C$ to steam at $100^{\circ} C$

1

3200 c a l

2

3100 c a l

3

4200 c a l

4

3600 c a l

Explanation:

Ice $(0^{\circ} C)$ converts into steam $(100^{\circ} C)$ in three steps.
The heat required $Q = Q_{1} + Q_{2} + Q_{3}$
$= 5 \times 80 + 5 \times 1 \times (100 - 0) + 5 \times 540 = 3600 c a l$

PHXI11:THERMAL PROPERTIES OF MATTER

366287 In similar calorimeters, equal volumes of water and alcohol, when poured, take $100 \sec$ and 74 sec respectively, to cool from $50^{\circ} C$ to $40^{\circ} C$ . If the thermal capacity of each calorimeter is numerically equal to volume of either liquid, then calculate the specific heat capacity of alcohol (given: the relative density of alcohol as 0.8 and specific heat capacity of water as
1 cal g $^{- 10} C^{- 1})$ .

1

0.8 c a l / g m^{\circ} C

2

0.6 c a l / g m^{\circ} C

3

0.9 c a l / g m^{\circ} C

4

1 c a l / g m^{\circ} C

Explanation:

Let $V$ be volume of either liquid
Mass of water $= V \times 1 g m$
Mass of alcohol $= V \times 0.8 = 0.8 V g m$
Rate of cooling of the water calorimeter
$= \frac{1}{100} [V \times (50^{\circ} - 40^{\circ}) + V \times (50^{\circ} - 40^{\circ})]$
$= (1 / 5) V c a l / \sec$
Rate of cooling of alcohol calorimeter $= \frac{1}{74} [V \times [50^{\circ} - 40^{\circ}] + 0.8 V \times c \times (50^{\circ} - 40^{\circ})]]$
$= (1 / 74) (10 V + 8 V c) c a l / \sec$
It is given that calorimetrers are similar to as rate of cooling of both is same
$\frac{1}{5} V = (1 / 74) 10 V + 8 V c$
$c = 0.6 c a l / g m^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366288 A chef, on finding his stove out of order, decides to boil coffee by shaking it in a thermosflask. Suppose that he uses water at $15^{\circ} C$ and that the water falls $30 c m$ each shake, the chef making 30 shakes each minute. Neglecting any loss of thermal energy by the flask, how long must he shake the flask until the water reaches $100^{\circ} C (c_{c o f f e e} = 4190 J)$ ?

1

2.25 \times 10^{3} min

2

3.97 \times 10^{3} min

3

4.03 \times 10^{3} min

4

5.25 \times 10^{3} min

Explanation:

The time taken to boil the coffee is
Hence, $t = \frac{Q}{N m g h} = \frac{c m (T_{f} - T_{i})}{N m g h} = \frac{c (T_{f} - T_{i})}{N g h}$
$= \frac{(4190) (100 - 15)}{(30) (9.8) (0.30)} = 4.03 \times 10^{3} \min$

PHXI11:THERMAL PROPERTIES OF MATTER

366289 Which of the following states of matter have two specific heats?

1 Solid

2 Gas

3 Liquid

4 All three

Explanation:

For solids and liquids $C_{P} \approx C_{V}$ . Whereas for gases $C_{P} - C_{V} = R$

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