Which of the following statements about the thermodynamic scale is not correct?
- It is also known as the Kelvin scale.
- It is a shifted scale from the Celsius scale by a value of 273.16.
- One degree change in the thermodynamic scale is equivalent to one degree change in the Celsius scale.
- The zero value of the thermodynamic scale represents the minimum internal energy of all substances.
The value should be 273.15, not 273.16.
A fixed mass of an ideal gas loses 2190 J of heat and expands under a constant pressure of 11 kPa from a volume of 25 x 10-3 m3 to a volume of 50 x 10-3 m3.
What is the change in the internal energy of the gas?
- -1920 J
- -2470 J
- 1920 J
- 2470 J
From the first law of thermodynamics,
ΔU = Q + W
Q = -2190 J
W = – P(Vi – VF
= -11 x 103 (50 x 10-3 – 25 x 10-3)
= -275 J
ΔU = -2190 – 275
= -2465 J
= -2470 J (corrected to 3 s.f.)
Why is the specific latent heat of fusion of any substance always less than its specific latent heat of vaporization?
In the melting process, molecules need only to break down the structure into a less-ordered arrangement of molecules (bonds are still present between the molecules, they are just weakened). In vaporisation however, the bonds are completely broken and this process requires a greater amount of energy than melting requires.
As a solid melts into a liquid, its volume does not increase substantially and very little energy is required to do work against the atmospher as a result of this small volume increase. As a liquid vaporises into a gas, the increase in volume is very much larger and a lot more energy is required to do work against the atmosphere due to the large expansion. Hence a greater amount of energy is required for vaporisation compared to fusion.
A flask contains 300 g of water at 80°C. Initially, the flask is at thermal equilibrium with the water. 40 g of ice at 0°C is added to the water in the flask. After some time, when all the ice had melted, the temperature of the flask and its contents was found to be 50°C.
a) Neglecting any heat lost to the surroundings, show that
Qflask = 0.040lf – 29400, where
Qflask is the heat lost by the flask and lf is the specific latent heat of fusion of ice.
(Specific heat capacity of water = 4200 J kg-1K-1
b) A further 65 g of ice was then added to the flask. After all the ice had melted, it was noted that the temperature of the flask and its contents was 20°C.
Q’flask, the heat lost by the flask after 65 g of ice was added was found to be related to lf, the specific latent heat of fusion of ice by the expression
Q’flask = 0.065lf – 37380
i) Explain why the heat lost by the flask is the same for both situations in a) and b).
ii) Hence, find lf, the specific latent heat of fusion of ice.
(a)Heat gained by the ice for melting = 0.04 lf
Heat gained by the melted ice to raise its temperature from 0°C to 40°C = 0.04 x 4200 x (50 – 0)
Heat lost by the water in the flask = 0.3 x 4200 x (80 – 50)
Heat lost by the flask = mf x cf x (80 – 50), where mf and cf is the mass and specific heat capacity of the flask respectively.Heat gained = Heat lost
0.04 lf + 0.04 x 4200 x (50 – 0) = 0.3 x 4200 x (80 – 50)+ mf x cf x (80 – 50)
Qf = mf x cf x (80 – 50) – 0.3 x 4200 x (80 – 50)
Qf = 0.04lf – 29400 J
i) The temperature change is the same for both situations, hence the heat lost by the flask will be the same.
0.040lf – 29400 = 0.065lf – 37380
lf = 3.19 x 105 J kg-1
In an experiment with a continuous flow calorimeter to find the specific heat capacity of a liquid, an input power of 60 W produced a rise in temperature of 10 K in the liquid. When the power was doubled, the same temperature rise was achieved by making the rate of flow of the liquid three times faster.
The power lost to the surroundings in each case was
- 20 W
- 30 W
- 40 W
- 50 W
E = mcΔθ + heat lost
Pt = mcΔθ + heat lost
60t = mcΔθ + heat lost —- eqn 1
120t = 3mcΔθ + heat lost —- eqn 2 (3m because rate of flow of liquid is 3 times faster)
Solving eqn 1 and eqn 2,
heat lost = 30W