**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.**

**Show/Hide Answer**

The value should be 273.15, not 273.16.

Answer: 2

**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} m^{3} to a volume of 50 x 10^{-3} m^{3}.**

**What is the change in the internal energy of the gas?**

**-1920 J****-2470 J****1920 J****2470 J**

**Show/Hide Answer**

From the first law of thermodynamics,

ΔU = Q + W

Q = -2190 J

W = – P(V

_{i}– V

_{F}

= -11 x 10

^{3}(50 x 10

^{-3}– 25 x 10

^{-3})

= -275 J

ΔU = -2190 – 275

= -2465 J

= -2470 J (corrected to 3 s.f.)

Answer: 2

**Why is the specific latent heat of fusion of any substance always less than its specific latent heat of vaporization?**

**Show/Hide Answer**

Answer 1:

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.

Answer 2:

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**

**Q _{flask} = 0.040l_{f} – 29400, where**

**Q**

_{flask}is the heat lost by the flask and*l*is the specific latent heat of fusion of ice._{f}**(Specific heat capacity of water = 4200 J kg ^{-1}K^{-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 l_{f}, the specific latent heat of fusion of ice by the expression**

**Q’ _{flask} = 0.065l_{f} – 37380**

**i) Explain why the heat lost by the flask is the same for both situations in a) and b).**

**ii) Hence, find l_{f}, the specific latent heat of fusion of ice.**

**Show/Hide Answer**

(a)Heat gained by the ice for melting = 0.04

*l*

_{f}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 = m

_{f}x c

_{f}x (80 – 50), where m

_{f}and c

_{f}is the mass and specific heat capacity of the flask respectively.Heat gained = Heat lost

0.04

*l*+ 0.04 x 4200 x (50 – 0) = 0.3 x 4200 x (80 – 50)+ m

_{f}_{f}x c

_{f}x (80 – 50)

Q

_{f}= m

_{f}x c

_{f}x (80 – 50) – 0.3 x 4200 x (80 – 50)

Q

_{f}= 0.04

*l*– 29400 J

_{f}b)

i) The temperature change is the same for both situations, hence the heat lost by the flask will be the same.

ii)

0.040*l _{f}* – 29400 = 0.065

*l*– 37380

_{f}*l*= 3.19 x 10

_{f}^{5}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**

**Show/Hide Answer**

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

Answer: 2

Hello again. Same formatting problem here. No criticism just pointing it out like you’ve asked.

Thanks! I’ve fixed the problem.