O lvl Phy: Thermal Properties of Matter

A mercury-in-glass thermometer is calibrated. The lengths of the mercury thread at the upper and lower fixed points (100°C and 0°C) are 24.0 cm and 4.0 cm respectively.

(a) The thermometer is then immersed in a beaker of water at room temperature. It is found that the length of the mercury thread is 9.0 cm. Find the room temperature.

(b) A metal cube at 80°C is immersed into the beaker of water. The thermometer in (a) is then immersed in the water. The length of mercury thread keeps increasing until it becomes steady at 9.4 cm.
(i) What is the final temperature of the water-metal system?
(ii) Given that the mass of water is 1 kg and the specific heat capacity of water is 4200 J kg^-1 K^-1, calculate the heat capacity of the metal.

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Answer:

(a) t/100 = (9 - 4) / (24 - 4)
t = 5 / 20 * 100
t = 25 °C

(b)
(i) t/100 = (9.4 - 4) / (24 - 4)
t = 5.4 / 20 * 100
t = 27 °C

(ii) Heat gained by water = heat loss by metal
Using Q = mcΔθ for water, and Q = CΔθ for the metal (c is specific heat capacity, C is heat capacity),
(m_water) (c_water) Δθ_water = (C_metal) Δθ_metal
(1)(4200)(2) = (C_metal) (80 - 27)
C_metal = 8400 / 53
C_metal = 158.5 J K^-1

O lvl Phy: Transfer of Heat Energy, Thermal Properties of Matter

RGS Prelims 1997 P2 Q10

(a) The figure below shows an experiment on heat transfer.

(i) State the means of heat transfer from the heater to thermometers X and Y.
(ii) Which thermometer will have a higher temperature change after a certain time?
(iii) If the heater is placed between thermometers X and Y, determine which one will register a higher change. Give reasons.

(b) A 3 kW immersion heater is used to keep the water in a domestic hot water tank at a steady temperature. The electrical supply to the immersion heater is switched off when the average water temperature is 60 °C. Assuming that the rate of heat loss is constant at 1.8 kW and no water is run off, and that the total heat capacity of the tank and the water it contains is 5.2 * 10⁵ J K^-1, calculate the average temperature 40 minutes later.

(c) Liquid air boils at a very low temperature at normal atmospheric pressure. Explain why liquid air contained in an open vacuum flask in a laboratory boils steadily and continuously. Why does liquid air boil much more rapidly when contained in an ordinary glass beaker?

(d) A 20 g lump of iron is placed in liquid air for several minutes. It is then removed and quickly placed in water at 0 °C. A 5.2 g layer of ice forms over the iron. Determine the temperature of the liquid air. (The specific heat capacity of iron is 440 J kg^-1 °C^-1 and the latent heat of fusion of ice is 3.34 * 10⁵ J kg^-1.)

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Answer:

(a)
(i) Conduction (Not convection because heater is on top, and hot water rises)
(ii) Thermometer X (nearer to heater)
(iii) Thermometer X. This is because hot water rises and cold water sinks. The water above the heater (where thermometer X is) will get heated up faster.

(b) Using Q = CΔӨ,
1.8 * 1000 * (40 * 60) = 5.2 * 10⁵ * ΔӨ
ΔӨ = (4.32 * 10⁶) / (5.2 * 10⁵)
ΔӨ = 8.3 °C

Hence, average temperature = 60 °C - 8.3 °C = 51.7 °C

(c) Liquid air contained in an open vacuum flask in a laboratory boils steadily and continuously because the room temperature is always higher than the boiling point of liquid air; heat energy is being constantly supplied from the surroundings to the liquid air to boil it continuously.

Liquid air boils much more rapidly when contained in an ordinary glass beaker because in an ordinary glass beaker, heat energy is also supplied through conduction via the sides of the beaker. Heat transfer to liquid air in a vacuum flask is limited to conduction via the hole/entrance of the vacuum flask. Thus, we can see that heat energy is being supplied at a higher rate for the glass beaker as compared to the vacuum flask (more channels/surface area for heat transfer). This explains why liquid air boils much more rapidly when contained in an ordinary glass beaker.


(d) Let the temperature of liquid air be T
Heat gained by Iron = Latent heat loss to change from water at 0 °C to ice at 0 °C.
mc_iron ΔӨ = ml_fusion
(0.020)(440)(0 - T) = (0.0052)(3.34 * 10⁵)
-8.8T = 1736.8
T = -197 °C

Note: A quick way to check... Boiling point of liquid nitrogen is around -196 °C, so since air is mainly nitrogen, -197 °C sounds very logical.

O lvl Phy: Work Energy Power

RI 1998 Sec 3 EOY P2 Q7

A substance of mass 0.30 kg initially in the solid state, is heated by a 100 W heater. The figure below shows how the temperature of the substance varies with time.

(a) Determine the melting point of the substance.

(b) Calculate
(i) the specific heat capacity of the substance in the liquid state.
(ii) the specific latent heat of fusion of the substance.

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Answer:

(a) From the figure, the melting point of the substance is 80 °C


(b)
(i) Using Q = mcΔθ,
Q = 100 W * 100 s = 10000 J
m = 0.30 kg
Δθ = 100 - 80 = 20

Hence, specific heat capacity, c = Q / (mΔθ)
= 10000 / (0.30 * 20)
= 1.7 * 10³ J kg^-1 K^-1

(ii) Using Q = ml,
Q = 100 W * (1050 s - 300 s) = 75000 J
m = 0.30 kg

Hence, specific latent heat, l = Q / m
= 75000 / 0.30
= 2.5 * 10⁵ J / kg

O lvl Phy: Thermal Properties of Matter

Some crushed ice, at 0 °C is dried quickly with a blotting paper and transferred to a known mass of water in an insulated container.

The following results were obtained.

Initial mass of water in beaker at the start = 100g
Mass of ice transferred to the beaker = 10g
Initial temperature of water = 23 °C
Final temperature of water = 16 °C

(a) Given that water has a specific heat capacity of 4200 J/(kg °C), calculate
(i) the heat energy released by the water in cooling,
(ii) the latent heat to melt the ice,
(iii) a value for the specific latent heat of fusion for ice.

(b) Would the value obtained for (a)(iii) likely to be higher or lower than the actual value? Explain your answer.

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Answer:

(a)
(i) Heat energy released by water = mcΔθ = 0.1 * 4200 * (23 - 16) = 2940 J

(ii) Heat energy released by water = latent heat to melt the ice + heat energy gained by water (from melted ice)
2940 = latent heat to melt the ice + 0.01 * 4200 * (16 - 0)
2940 = latent heat to melt the ice + 672
latent heat to melt the ice = 2940 - 672 = 2268 J

(iii) Let the specific latent heat of fusion for ice be L

latent heat to melt the ice = mL = 0.01 L
2268 = 0.01 L
L = 226800 J / kg


(b) It would be higher.

This is because heat is also released by the beaker,

i.e.Heat energy released by water + Heat released by beaker = latent heat to melt the ice + heat energy gained by water (from melted ice)

Thus, latent heat to melt the ice = Heat energy released by water + Heat released by beaker - heat energy gained by water (from melted ice)



Note: You cannot say heat is gained (or lost) to the surroundings, because it is already stated that the water is in an insulated container.

O lvl Phy: Thermal Properties of Matter

A glass contains 650 g of water at 25 °C. 150 g of ice at -5 °C is put into the water. Calculate the final temperature of the contents.

Specific heat capacity of ice = 2030 J/kg K
Latent heat of fusion of water = 3.36 * 10⁵ J/kg
Specific heat capacity of water = 4200 J/kg K
Latent heat of vaporisation of water = 2.26 * 10⁶ J/kg

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Answer:

Energy required for ice from -5 °C to 0 °C
Q = mcΔθ = 0.15 * 2030 * 5
= 1522.5 J

Energy required for ice to convert from ice at 0 °C to water at °C
Q = ml = 0.15 * 336000
= 50400 J

Thus, total heat gained by ice = 51922.5 J

Note: This heat gained by ice (heat loss by the water) is not yet sufficient to bring the water to 0 °C and convert it to ice. Hence, we work assuming that the final temperature is somewhere between 25 °C and 0 °C, which we shall see.


Let the final temperature be t
total heat loss by water = total heat gained by ice
mc_waterΔθ (from water at 25 °C to water at t °C) = 51922.5 + mc_waterΔθ (from water at 0 °C to water at t °C)
0.65 * 4200 * (25 - t) = 51922.5 + 0.15 * 4200 * t
68250 - 2730 t = 51922.5 + 630 t
3360 t = 16327.5
t = 4.86 °C

O lvl Phy: Thermal Properties of Matter

The specific latent heat of fusion of an object is 2.5 times smaller than its specific heat capacity. 250g of the solid object is heated from its freezing point to 20 K above its freezing point by a 125 W heater applied for 8 minutes. Calculate both the specific latent heat of fusion and the specific heat capacity of the object.

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Answer:

Let the specific latent heat of fusion = l
thus, the specific heat capacity = 2.5 l

Total energy = Pt = ml + mcΔθ
125 * 8 * 60 = 0.25 * l + 0.25 * 2.5 l * 20
60000 = 0.25 l + 12.5 l
12.75 l = 60000

Thus, specific latent heat of fusion, l = 4705.9 J/kg

Specific heat capacity = 2.5 l = 11764.7 J /kg K

O lvl Phy: Thermal Properties of Matter

A heater of power, P watts, supplies thermal energy to 100g of ice at -8 °C until it fully converts to steam at 100 °C. This heating process took 5.5 minutes. Calculate the value of P.

Specific heat capacity of ice = 2030 J/kg K
Latent heat of fusion of water = 3.36 * 10⁵ J/kg
Specific heat capacity of water = 4200 J/kg K
Latent heat of vaporisation of water = 2.26 * 10⁶ J/kg

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Answer:

Total energy = Pt = ( energy to change ice at -8 °C to ice at 0 °C ) + ( energy to change from 100g of ice to 100g of water ) + ( energy to change from 100g of water at 0 °C to 100g of water at 100 °C ) + ( energy to change from 100g of water to 100g of steam)

Pt = (mcΔθ)_ice + (ml)_ice + (mcΔθ)_water + (ml)_water
P * 5.5 * 60 = (0.1 * 2030 * 8) + (0.1 * 336000) + (0.1 * 4200 * 100) + (0.1 * 2260000)
330P = 1624 + 33600 + 42000 + 226000
P = 918.86 W