[2 Mark Question Answer]

Question 512 Marks

In an experiment. 17 g of ice is used to bring down the temp. of 40 g of water from 34°C to its freezing point. The sp. heat capacity of water is 4.25 J/g°C. Calculate sp. latent heat of ice.

Answer

Heat gained by ice at 0°C to convert to water at 0°C = Heat lost by water from 34°C to 0°C
17 x L = 40 x 4.25 x 34
17 L = 5780
L = 340 J/g

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Question 522 Marks

40 g of ice at -16°C is dropped into water at 0°C, when 4 g of water freezes into ice. If specific heat capacity of ice is 2100 J/kg°C, what will be the latent heat of fusion of ice?

Answer

Let the latent heat of fusion of ice be L.
Heat gained by ice at -16°C to convert to 0°C = Heat given out by 4 g of water to at 0°C to freeze into ice at 0°C
(40 x 2.1 x -16) = 4 x L
1344 = 4 L
L = 336 J/g

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Question 532 Marks

(i) A molten metal weighing 150 g is kept at its melting point 800°C. When it is allowed to solidify at the same temp, it gives out 75000 J of heat. What is the specific heat of the metal?
(ii) If its specific heat capacity is 200 J/kgK? How much additional heat will it give out in cooling to -50°C?

Answer

Let the specific latent heat of metal is $L$.
(i) When the metal is allowed to solidify at its m.p./ i.e., $800^{\circ} \mathrm{C}$, it will give out its latent heat only.
Therefore $\mathrm{mL}=75000$
or $0.150 \times \mathrm{L}=75000$
$ \therefore \mathrm{L}=\frac{75000}{1150}=65217.3 \mathrm{~J} / \mathrm{Kg} \text {. } $
(ii) Heat given out by the metal piece in cooling down to $-50^{\circ} \mathrm{C}$. Fall in temperature of the metal piece
$ \begin{aligned} & =[800-(-50)]=850^{\circ} \\ & \therefore \text { Heat given out }=\mathrm{m} \times \mathrm{s} \times \mathrm{t}=(0.150 \times 200 \times 850) \mathrm{J} \\ & =25500 \text { joule } \end{aligned} $

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Question 542 Marks

Define the following terms:
(i) Specific latent heat,
(ii) Specific latent heat of fusion.

Answer

(i) sSpecific latent heat is the amount of heat required to change the state of unit mass of a substance without in temperature.
(ii) Specific latent heat of fusion isthe amount of heat required to change unit mass of a solid at its melting liqud at the same temperature.

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Question 552 Marks

Define the following terms:
(i) Latent heat,
(ii) Latent heat of fusion of ice.

Answer

(i) Latent heat is the amount of hidden heat supplied or extracted from the substance to change its state change of temperature.
(ii) Latent heat of fusion of ice is the amount heat absorbed by ice at 0°C to convert into water at 0°C.

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Question 562 Marks

A mass of 200 g of mercury at 100°C is poured into water at 20°C. If the final temperature of the mixture is 25°C, find the mass of water. (Sp. heat cap. of mercury is 140 J/kgK).

Answer

Let the mass of water be $m$.
Heat lost by mercury $=$ Heat gained by water
$0.2 \times 140 \times(100-25)=m \times 4200 \times(25-20)$
or, $\mathrm{m}=\frac{0.2 \times 140 \times 75}{4200 \times 5}=0.1 \mathrm{~kg}=100 \mathrm{~g}$

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Question 572 Marks

A solid of mass 80 g at 80°C is dropped in 400 g water at 10°C. If final temp. is 30°C, find the sp. heat cap. of the solid.

Answer

Let the specific heat of the solid be $\mathrm{c}$.
Heat lost by solid = Heat gained by water
$ 0.08 \times \mathrm{c} \times(80-30)=0.4 \times 4200 \times(30-10) $
or, $C={ }^{\prime}(0.4 x x 4200 x x 20) /(0.08 x x 50)=8400 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

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Question 582 Marks

400 g of water at 80°C is mixed with 1 kg of water at 20°C. If the sp. heat cap. of water is 4200 J/kgK, find the final temperature of the mixture.

Answer

Let the final temperature of the mixture be 8°C.
Heat lost by hot water = Heat gained by oold water
0.4 x C x (80 - θ) = 1 x C x (θ - 20)
or, 32 - 0.48 = θ - 20
or, θ = 37.14°C

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Question 592 Marks

Calculate the amount of heat given out by 600 g of aluminium while cooling from $100^{\circ} C$ to $30^{\circ} C$. ( Sp . heat cap. of aluminium $=900 J / kgK$.)

Answer

Change in temperature $=(100-30)=70^{\circ} C =70 K$
Amount of heat given out, $Q = m \times C \times T$
$=0.6 \times 900 \times 70=37800 J$

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Question 602 Marks

What amount of heat would be required to raise the temperature of 500 g of water from 30°C to 50°C? (sp. heat cap. of water= 4200 J/kg°C)

Answer

Change in temperature = (50-30) = 20°C
Amount of heat required, Q = m x C x ?T
= 0.5 x 4200 x 20 = 42000 J

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Question 612 Marks

A slab of ice at -5°C is constantly heated till it changes to steam at 105°C. Draw a graph showing the change of temperature with time and label the different parts of the graph.

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Question 622 Marks

Define heat capacity and state its units.

Answer

Heat capacity of a body is the quantity of heat required to raise its temperature by 1°C. It depends upon the mass and the nature of the body.
Units: J/°C or calorie/°C

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Physics [2 Mark Question Answer]