A101 ScienceProblem 13: Cool it!

6th Presentation

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• We know that Jonathan is supposed to get a final sample of water at 15 °C by mixing ice with tap-water.

• We also know that the ice will first melt into its liquid state, followed by a temperature increase.

• At the same time, the temperature of the tap-water will decrease.

• Assuming we use 1 kg of tap-water at 25 °C and a certain amount of ice at 0 °C.

• Jonathan must add the correct amount of ice so that both melted ice and tap-water end up as water at exactly 15 °C.

What we know about the problem

• Heat is a form of energy while temperature is a measure of how hot or cold a substance is.

• Whenever a substance at a higher temperature is brought into contact with a substance at a lower temperature, heat will flow from the substance with the higher temperature to the substance with the lower temperature.

Heat and temperature

Substanceat

0 °C

Substance at

100 °C

Heat flows from the hottersubstance to the cooler substance It is assumed that there is no heat

gain or loss to the surroundings

• When a substance loses heat, its temperature will decrease.

• The opposite is also true, that is when a substance gains heat, its temperature will increase.

• Heat flows such that the two substances reach the same temperature.

Heat and temperature

Substance at

highertemperature

Substanceat

lowertemperature

Direction of heat flow

Temperature decreasesas the substance loses heat

Temperature increasesas the substance gains heat

It is assumed that there is no heat gain or loss to the surroundings

Substance reaches a final temperaturelower than its original temperature

Substance reaches a final temperaturehigher than its original temperature

Substances at thesame

temperature

• When ice is added to tap-water, the following occurs- Tap-water loses heat to the ice and its temperature drops.- Ice gains heat from the tap-water and starts to melt at 0 °C.

• After all the ice has melted at 0 °C,- Tap-water continues to lose heat to the melted ice and its

temperature drops further.- Melted ice gains heat from the tap-water and its temperature

increases.

• The above temperature change occurs till both the temperature of tap-water and melted ice are equal.

Interaction between ice and water

Mass(m)/kg

Heat Supplied (Q)/

kJ

Initial Temperature/

°C

Final Temperature/

°C

Change in Temperature* (∆T)/

K1 10 25 30 51 20 25 35 101 30 25 40 151 40 25 45 201 10n 25 (25 + 5n) 5n

• It can be deduced that when heat supplied is n times of its original value, the temperature change must also be n times of its original value as well.

• Hence the heat supplied is directly proportional to the temperature change.

Relationship between heat supplied (Q) and temperature change (∆T)

Heat suppliedis doubled

Change in temperatureis doubled

Heat suppliedis halved

Change in temperatureis halved

Heat supplied is n times

Change intemperature is n times*Change in 1 K = Change in 1 °C

Mass(m)/kg

Heat Supplied (Q)/

kJ

Initial Temperature/

°C

Final Temperature/

°C

Change in Temperature

(∆T)/K1 10 25 30 5

2 20 25 30 5

3 30 25 30 5

4 40 25 30 5

n 10n 25 30 5

• It can be deduced that when mass is n times of its original value, the amount of heat supplied must also be n times of its original value.

• Hence the heat supplied is directly proportional to the mass.

Relationship between heat supplied (Q) and mass (m)

Mass isdoubled

Heat suppliedis doubled

Mass ishalved

Heat suppliedis halved

Massis n times

Heat supplied is n times

Combining the different relationships established in the previous slides, we can arrive at the relationship which links the heat supplied (Q), the amount of a substance (m) and the change in temperature (∆T):

Q = m × c × ∆T

where c is the proportionality constant known as the specific heat capacity.

Relationship between heat supplied (Q), mass (m) and temperature change (∆T)

• By considering the ratio of heat supplied (Q) to the product of mass multiplied by change in temperature (m x ∆T), we find that this ratio is a constant for each type of substance.

• This ratio is defined as the specific heat capacity of a substance.

• The specific heat capacity represents the fixed amount of heat which is required to change the temperature of 1 kg of a given substance by 1 K (e.g. from 50 °C to 51 °C).

• The Kelvin (K) scale is a commonly used temperature scale in which a change in temperature of 1 K is the same as a change in temperature of 1 °C.

Specific heat capacity

• The value of specific heat capacity is different for different substances.

• In the case of water, 4.184 kJ of heat energy is needed to raise the temperature of 1 kg of water by 1 K.Specific heat capacity of water = 4.184 kJ per kg per Kwhich is more commonly written as 4.184 kJ/(K∙kg)

Specific heat capacity

Mass (m)/kg

Heat Supplied

(Q)/kJ

Initial Temperature/

°C

Final Temperature/

°C

Change In Temperature

(∆T)/K kJ/(K∙kg)

1 10 25 27.390 2.390 4.184

1 20 25 29.780 4.780 4.1841 30 25 32.170 7.170 4.1841 40 25 34.560 9.560 4.184

• The higher the specific heat capacity, the greater the amount of heat which must be supplied to raise the temperature of a given mass of the substance.This also means that a greater the amount of heat needs to be removed so as to bring about a reduction in temperature.

• For example, 2 kg of Substance X (with a higher specific heat capacity of 2.5 kJ/(K∙kg) as compared to 2 kg of Substance Y (with a lower specific heat capacity of 2 kJ/(K∙kg) requires more heat to be removed to reduce its temperature by the same amount.

Specific heat capacity

• The fixed amount of heat which is required to change 1 kg of a given substance from the one physical state into another physical state without a change in its temperature is known as the specific latent heat.

• The specific latent heat involved in the melting or freezing of a substance is called specific latent heat of fusion.

• The specific latent heat involved in the boiling or condensation of a substance is called specific latent heat of vaporisation.

• The values of specific latent heats are different for different substances.

Specific latent heat

• In the case of ice, 334 kJ of heat energy is needed to convert 1 kg of ice at 0 °C into 1 kg of melted ice at 0 °C.

Specific latent heat of fusion of ice = 334 kJ per kg which is more commonly written as 334 kJ/kg

• In order to convert 2 kg of ice at 0 °C into 2 kg of melted ice at 0 °C, (2 x 334) kJ = 668 kJ of heat is absorbed by the ice.

• Similarly, to convert 3 kg of ice at 0 °C into 3 kg of melted ice at 0 °C, (3 x 334) kJ = 1002 kJ of heat is needed.

Specific latent heat of fusion

• Therefore, we can write an useful relationship which links the heat supplied (Q), the amount of a substance (m) and the specific latent heat of fusion (L) as

Q = m × L• The higher the specific latent heat of fusion, the greater

the amount of heat which must be supplied to melt a given mass of the substance.

Specific latent heat of fusion

• In general, when heat is constantly added to a solid, its Temperature vs. Heat Supplied profile can be represented in the graph on the right

• As heat is continuously supplied to a solid, the substance progresses from Phase A to E.

Heating a Solid

Heat Supplied / kJ

Temperature / °C

A

C

B

D

E

Phase State Temperature Process Property Involved Equation

A Solid Increases - Specific Heat Capacity of Solid, cs Q = m x cs x ∆T

B Solid & Liquid Stays Constant Melting Specific Latent Heat of Fusion, Lf Q = m x Lf

C Liquid Increases - Specific Heat Capacity of Liquid, cl Q = m x cl x ∆T

D Liquid & Gaseous Stays Constant BoilingSpecific Latent Heat of

Vaporization, LvQ = m x Lv

E Gaseous Increases - Specific Heat Capacity of Gas， cg Q = m x cg x ∆T

Problem Solving: Cooling tap-water down• We know that in the case of tap-water, 4.184 kJ of heat

energy is needed to raise the temperature of 1 kg of tap-water by 1 K (e.g. from 50 °C to 51 °C).

• We must remove the same amount of heat in order to reduce the temperature of 1 kg of tap-water by 1 K (e.g. from 51 °C to 50 °C).

• Therefore, we can deduce that Jonathan needs to remove 41.84 kJ of heat in order to cool 1 kg of tap-water from 25 °C to 15 °C.

Mass (m)/ kg

Heat Supplied (Q)/kJ

Initial Temperature /C

Final Temperature /C

Change in Temperature

(∆T)/K

1 -4.184 51 50 -1

1 25 15 -10

Change in temperatureis 10 times!

Heat to be removedis 10 times!

-41.84

• Heating ice to water at a certain temperature requires 1) the melting of ice to melted ice at 0 °C and 2) increasing the temperature of melted ice from 0 °C to the desired temperature.

1) Melting 1 kg of ice at 0 °C into 1 kg of melted ice at 0 °C requires 334 kJ of heat.Therefore, (m × 334) kJ of heat is needed to turn m kg of ice at 0 °C into m kg of the melted ice at 0 °C.

2) To increase the temperature of melted ice at 0 °C to 15 °C, a further (m × 4.184 × 15) kJ of heat is needed.

Problem Solving: Heating ice up

Mass (m)/kg

Heat Supplied (Q)/

kJ

Initial Temperature/

C

Final Temperature/

C

Change in Temperature (∆T)/

K

1 4.184 0 1 1

0 1 1

0 15 15m (m × 4.184)

Mass increases by m times

Temperature change is constant

Heat suppliedmust increase by m times

m (m × 4.184 × 15)

Mass kept constant

Temperature change increases by 15 times

Heat supplied mustincrease by 15 times

• In order for Jonathan to obtain water at 15 °C from an initial temperature of 25 °C,

Qlost by tap-water = Qgained by ice to melt + Qgained by melted ice

(1 × 4.184 × 10) kJ = (m × 334) kJ + (m × 4.184 × 15) kJ

m = 0.105 kg

• Therefore, for every 1 kg of tap-water at room temperature to be cooled, 0.105 kg of ice is needed.

Combining the two processes

• Land heats up faster than water in the day, and cools down faster in the night because land has a lower specific heat capacity than water.

• During the day, City A is surrounded by hot land while City B is surrounded by cooler water.

• City A will have a higher temperature during the day.

Going further

• At night, City A is surrounded by cooler land while City B is surrounded by warmer water.

• City A will have a lower temperature during the night.• The daily temperature difference of City A is much larger

than that of City B.

Learning points • Heat is a form of energy while temperature is a measure of

how hot or cold a substance is.• Heat always flows from a region of higher temperature to a

region of lower temperature when two substances with different temperatures are brought directly in contact with each other.

• Specific heat capacity of a substance is the fixed amount of heat needed to change the temperature of 1 kg of that substance by 1 K.

• Specific latent heat of a substance is the fixed amount of heat needed to change 1 kg of that substance from the one state into another state without a change in its temperature.

For a faster response, should we use a fluid with a high or low specific heat capacity when making a thermometer? (Assume that all other factors remain constant) Explain your answer.

Discussion