Physics [3 Mark Question Answer]

The molecules in a solid are held by strong intermolecular bonds. For the solid to melt, these bonds have to be broken. Since energy is needed to break the intermolecular bonds, the thermal energy supplied at the melting point is used to do the work to break the intermolecular bonds between the molecules of the solid. Once the intermolecular bonds are broken, the molecules can then move out of their fixed positions. Hence it can then be said that the solid has melted, which is the change of state from solid to liquid. This explains why temperature remains constant during the melting phases.

(a)Boiling point of substance is 150°C (because the part BC represents condensation where the vapour changes into liquid without the change in temperature.
(b)DE represents freezing of the substance where the liquid changes into solid at a constant temperature of 100°C.
(c)Melting point is the temperature of the region DE where liquid changes into solid i.e., 100°C.

i.AB represents the change of state from solid to liquid i.e., AB represents melting of ice at 0°C.
ii.CD represents the change of state from liquid to vapour i.e., CD represents boiling of water at 100°C.
iii.The ice initially is in solid state at -10°C. On heating, its temperature rises to 0°C. It then takes some heat at 0°C to melt in water at 0°C which is its latent heat.

Amount of heat required to oonvert ice into steam is as given below:
(ice from -10°C to 0°C) = 0.1 x 2100 x 10 = 2100 J
(ice at 0°C to water at 0°C) = 0.1 x 336000 = 33600 J
(water from 0°C to 100°C) = 0.1 x 4200 x 100 = 42000 J
(water at 100°C to steam at 100°C) = 0.1 x 2260000 = 226000 J
Total amount of heat required = 2100 + 33600 + 42000 + 226000 = 30370 J

Mass of lead block, m = 250 g
Change in temperature, ΔT = 327°C - 27°C = 300°C = 300 K
C = 0.13 J/gK
Amount of heat required to raise the temperature to 327°C
Q = m x C ΔT
= 250 x 0.13 x 300 = 9750 J
Amount of heat required to completely melt the block upto its melting point
Q = m x L
= 250 x 26 = 6500 J

Amount of heat given out while converting water from 50°C to 0°C
= m x c x θ
= 100 x 4.2 x 50
= 21000 J
Amount of heat given out while converting water at 0°C to ice at 0°C = mL
= 100 x 330
= 33000 J
Amount of heat given out while converting ice from 0°C to 50°C
= 100 x 2.1 x 5
= 10500 J
Total heat required during the process = 64500 J

Amount of heat requird to convert ice from -10°C to 0°C = m x c x θ
= 40 x 2.1 x 10 = 840 J
Amount of heat required to convert ice at 0°C to water at 0°C = mL
= 40 x 330 = 13200 J
Amount of heat required to convert water at 0°C to water at 20°C
= 40 x 4.2 x 20 = 3360 J
Total heat required during the process = 17400 J