Quantity of Heat Energy
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Specific Heat Capacity
The amount of heat required to change the temperature of a body depends on its mass and on the change in temperature.
i. Quantity of heat is directly proportional to mass:
Q ∝ m … (i)
ii. Quantity of heat is directly proportional to change in temperature:
Q ∝ ΔT … (ii)
Combining (i) and (ii):
Q ∝ mΔT
Q = msΔT … (iii)
where s is the proportionality constant called the specific heat capacity.
From (iii):
s = Q/(mΔT)
SI unit: J kg−1 K−1 (equivalently J kg−1 °C−1 for a temperature interval).
1 cal = 4.2 J
4200 J kg−1 °C−1 = 1 cal g−1 °C−1
Heat Capacity or Thermal Capacity
Q = msΔT
If ΔT = 1°C or 1 K:
Q = ms
Thus, the heat capacity of a body is the product of its mass and specific heat capacity.
Principle of Calorimetry
When two bodies at different temperatures are placed in contact, the hotter body loses heat and the colder body gains heat. Heat exchange continues until both attain the same temperature.
If there is no heat exchange with the surroundings, then:
Determination of Specific Heat Capacity of a Solid by Method of Mixture
Let:
According to the principle of calorimetry:
Heat lost = Heat gained
msss(T1 − T) = mwsw(T − T2) + mcsc(T − T2)
msss(T1 − T) = (mwsw + mcsc)(T − T2)
ss = [(mwsw + mcsc)(T − T2)] / [ms(T1 − T)]
Thus, knowing the specific heat of water and the calorimeter material, the specific heat capacity of the solid can be determined.
Newton’s Law of Cooling
Let T be the temperature of the body and Ts be the temperature of the surroundings.
dQ/dt ∝ −(T − Ts)
dQ/dt = −k(T − Ts) … (i)
where k is a proportionality constant. The negative sign indicates that the temperature difference decreases with time.
Also:
Q = msΔT
dQ/dt = ms(dT/dt) … (ii)
Using (i) and (ii):
ms(dT/dt) = −k(T − Ts)
dT/(T − Ts) = −(k/ms)dt
Integrating:
log(T − Ts) = −(k/ms)t + C
The graph is a straight line.
Determination of Specific Heat Capacity of a Liquid by Method of Cooling
Let:
Heat lost by calorimeter A and water:
(mAsc + mwsw)(T1 − T2)
Rate of cooling of A:
[(mAsc + mwsw)(T1 − T2)]/t1
Heat lost by calorimeter B and liquid:
(mBsc + mlsl)(T1 − T2)
Rate of cooling of B:
[(mBsc + mlsl)(T1 − T2)]/t2
Since both are cooled under identical conditions, their rates of cooling are taken equal:
(mAsc + mwsw)/t1 = (mBsc + mlsl)/t2
sl = [(mAsc + mwsw)t2/(mlt1)] − (mBsc/ml)
Latent Heat
The heat required during change of phase depends on the mass:
Q ∝ m
Q = Lm
where L is the proportionality constant called latent heat.
SI unit: J kg−1.
Types of Latent Heat
Measurement of Latent Heat of Fusion by Method of Mixture
Let mc be mass of calorimeter, mw mass of water, mi mass of ice, sw specific heat of water, sc specific heat of calorimeter, Lf latent heat of fusion of ice, T1 initial temperature of water and calorimeter, and T2 final temperature of mixture.
Heat gained by ice:
miLf + miswT2
Heat lost by water and calorimeter:
(mcsc + mwsw)(T1 − T2)
By calorimetry:
miLf + miswT2 = (mcsc + mwsw)(T1 − T2)
Lf = [(mcsc + mwsw)(T1 − T2)]/mi − swT2
Determination of Latent Heat of Vaporization of Water
Let mc be mass of calorimeter, mw mass of water, ms mass of steam, sc specific heat of calorimeter, sw specific heat of water, Lv latent heat of vaporization, T1 initial temperature of water and calorimeter, and T2 final temperature of mixture.
Heat lost by steam:
msLv + mssw(100 − T2)
Heat gained by water and calorimeter:
(mcsc + mwsw)(T2 − T1)
By calorimetry:
(mcsc + mwsw)(T2 − T1) = msLv + mssw(100 − T2)
Lv = [(mcsc + mwsw)(T2 − T1)]/ms − sw(100 − T2)
Solved Numericals from the Scanned Notes
Numerical 1 – Specific Heat Capacity of Aluminium
Question: A copper calorimeter of mass 300 g contains 500 g of water at 15°C. A 560 g block of aluminium at 100°C is dropped into the calorimeter and the temperature rises to 22.5°C. Find the specific heat capacity of aluminium.
Heat lost by aluminium = Heat gained by water + calorimeter
0.56sAl(100 − 22.5) = 0.5(4200)(22.5 − 15) + 0.3(390)(22.5 − 15)
43.4sAl = 15750 + 877.5
sAl = 383.12 J kg−1 °C−1
Numerical 2 – Specific Heat Capacity of a Metal
Question: In an experiment, a 200 g block of metal at 150°C is dropped into a copper calorimeter containing 150 cm³ of water at 27°C. The final temperature is taken in the source calculation as 40°C. Calculate the specific heat of the metal.
Mass of water = 0.15 kg
By calorimetry:
mmsm(150 − 40) = mcsc(40 − 27) + mwsw(40 − 27)
The handwritten working proceeds to:
sm = 434.5 J kg−1 °C−1
This numerical is retained according to the handwritten values and final answer shown in the supplied scan.
Numerical 3 – Iron Block in Water and Copper Pot
Question: A copper pot of mass 0.5 kg contains 0.170 kg of water at 20°C. A 0.250 kg block of iron at 85°C is dropped into the pot. Find the final temperature, assuming no heat loss to the surroundings.
The source uses siron = 470 J kg−1 °C−1, swater = 4200 J kg−1 °C−1, and scopper = 390 J kg−1 °C−1.
0.25(470)(85 − T) = [0.17(4200) + 0.5(390)](T − 20)
117.5(85 − T) = 909(T − 20)
T ≈ 27.4°C
Repeated Source Question – Q.1(D)
Page 11 repeats the aluminium-calorimetry question and explicitly says to refer to Q.1(A). Its complete solution is therefore the same as Numerical 1 above.
Numerical 4 – Initial Temperature of a Copper Ball
Question: A copper ball weighing 400 g is transferred from a furnace to a copper calorimeter of mass 300 g containing 1 kg of water at 20°C. The water rises to 50°C. Find the initial temperature of the ball.
Heat lost by ball = Heat gained by calorimeter + water
0.4(390)(T − 50) = 0.3(390)(50 − 20) + 1(4200)(50 − 20)
T ≈ 880°C
Numerical 5 – Aluminium Block, Final Temperature 25°C
Question: A copper calorimeter of mass 300 g contains 500 g of water at 15°C. A 560 g aluminium ball at 100°C is dropped into it and the temperature rises to 25°C. Find the specific heat capacity of aluminium.
0.56s(100 − 25) = [0.5(4200) + 0.3(390)](25 − 15)
42s = 21000 + 1170
s = 527.85 J kg−1 °C−1
Numerical 6 – Convert 10 kg Ice at −10°C into Steam at 100°C
Question: How much heat is required to convert 10 kg of ice at −10°C into steam at 100°C?
sice = 2100 J kg−1 K−1, Lf = 3.36 × 105 J kg−1, Lv = 2.268 × 106 J kg−1
Q1 = msice[0 − (−10)] = 10 × 2100 × 10 = 2.1 × 105 J
Q2 = mLf = 10(3.36 × 105) = 3.36 × 106 J
Q3 = msw(100 − 0) = 10 × 4200 × 100 = 4.2 × 106 J
Q4 = mLv = 10(2.268 × 106) = 2.268 × 107 J
Q = Q1 + Q2 + Q3 + Q4 = 3.045 × 107 J
Numerical 7 – Convert 5 kg Ice at −10°C into Steam at 100°C
Q1 = 5 × 2100 × 10 = 1.05 × 105 J
Q2 = 5 × 3.36 × 105 = 1.68 × 106 J
Q3 = 5 × 4200 × 100 = 2.1 × 106 J
Q4 = 5 × 2.268 × 106 = 1.134 × 107 J
Total heat ≈ 1.52 × 107 J
Numerical 8 – 10 g Ice at −10°C to Steam at 100°C
The source solves this numerical in calories using sice = 0.5 cal g−1 °C−1, Lf = 80 cal g−1, sw = 1 cal g−1 °C−1, and Lv = 540 cal g−1.
Q1 = 10 × 0.5 × 10 = 50 cal
Q2 = 10 × 80 = 800 cal
Q3 = 10 × 1 × 100 = 1000 cal
Q4 = 10 × 540 = 5400 cal
Total = 7250 cal = 30450 J
Numerical 9 – 100 g Ice at 0°C Mixed with 100 g Water at 20°C
Question: Find the result of mixing 100 g of ice at 0°C with 100 g of water at 20°C in an iron vessel. The source takes the vessel mass as 100 g and specific heat as 0.1 cal g−1 °C−1.
An initial all-melting calculation gives a negative final temperature, showing that all the ice cannot melt.
At 0°C, heat lost by water and vessel:
Q = 100(1)(20) + 100(0.1)(20) = 2200 cal
If m grams of ice melt:
80m = 2200
m = 27.5 g
Final mixture: 127.5 g water + 72.5 g ice at 0°C.
Numerical 10 – 0.8 kg Ice at −10°C Mixed with 0.8 kg Water at 80°C
The initial calculation again shows that the ice cannot melt completely.
Heat lost by water cooling to 0°C:
0.8 × 4200 × 80 = 268800 J
Heat needed to warm 0.8 kg ice from −10°C to 0°C:
0.8 × 2100 × 10 = 16800 J
Heat remaining for fusion = 252000 J
mLf = 252000
m = 252000/336000 = 0.75 kg
Final mixture: 1.55 kg water + 0.05 kg ice at 0°C.
Numerical 11 – Steam Passed into Water and Ice
Question: 10 g of steam at 100°C is passed into a mixture of 100 g water and 10 g ice at 0°C. Find the resulting temperature.
Heat gained by ice and resulting water:
10(80) + 110(1)T = 800 + 110T
Heat lost by steam:
10(540) + 10(100 − T) = 6400 − 10T
800 + 110T = 6400 − 10T
120T = 5600
T = 46.67°C
Numerical 12 – Ice in Water with a Vessel
Question: Find the result of mixing 10 g ice at 0°C with 15 g water at 20°C in a vessel of mass 100 g and specific heat 0.09 cal g−1 °C−1.
An all-melting assumption produces an impossible negative temperature, so the final temperature is 0°C.
Heat lost by water and vessel to 0°C:
15(1)(20) + 100(0.09)(20) = 300 + 180 = 480 cal
Mass of ice melted = 480/80 = 6 g
Final mixture: 21 g water + 4 g ice at 0°C.
Numerical 13 – 100 g Ice at 0°C and 100 g Water at 100°C
Using Lf = 336 × 103 J kg−1 and sw = 4200 J kg−1 K−1:
0.1(4200)(100 − T) = 0.1(336000) + 0.1(4200)T
42000 − 420T = 33600 + 420T
T = 10°C
Numerical 14 – Newton’s Law of Cooling and Surrounding Temperature
Question: A substance takes 3 minutes to cool from 50°C to 45°C and 5 minutes to cool from 45°C to 40°C. Find the temperature of the surroundings.
Using the source’s cooling-rate comparison:
(50 − 45)/3 = −k(50 − Ts)
(45 − 40)/5 = −k(45 − Ts)
Dividing:
5/3 = (50 − Ts)/(45 − Ts)
Ts = 37.5°C
Numerical 15 – Steam Required to Heat a Water–Ice Mixture
Question: A mixture of 500 g water and 100 g ice at 0°C is kept in a copper calorimeter of mass 200 g. How much steam from a boiler must be passed into the mixture so that the temperature reaches 40°C?
Heat gained by water, calorimeter, melted ice, and the melted ice warming to 40°C:
[0.5(4200) + 0.2(390)]40 + 0.1(336000) + 0.1(4200)(40)
= 87120 + 33600 + 16800 = 137520 J
Heat lost per kg of steam:
2.268 × 106 + 4200(100 − 40) = 2.52 × 106 J kg−1
ms = 137520/(2.52 × 106)
ms = 0.05457 kg = 54.57 g
Numerical 16 – Ice at −6°C Dropped into Water at 0°C
Question: 50 g of ice at −6°C is dropped into water at 0°C. How many grams of water freeze? The source uses specific heat of ice = 2000 J kg−1 °C−1.
Heat needed to warm ice to 0°C:
Q = 0.05 × 2000 × 6 = 600 J
This heat is supplied by freezing water:
m(336000) = 600
m = 0.001785 kg = 1.785 g
Numerical 17 – Height Needed for Falling Ice to Melt Completely
Question: From what height should a block of ice be dropped so that it may melt completely?
Potential energy = latent heat required
mgh = mLf
Using g = 10 m s−2 and Lf = 336000 J kg−1:
10h = 336000
h = 33600 m
Numerical 18 – If Only 20% of Fall Energy is Retained by Ice
Question: From what height should a block of ice be dropped so that it may melt completely if only 20% of the energy of fall is retained by the ice?
0.20mgh = mLf
0.20 × 10 × h = 336000
h = 168000 m
Numerical 19 – Evaporation for Body Temperature Control
Question: Evaporation or perspiration is an important mechanism for temperature control of warm-blooded animals. What mass of water must evaporate from the surface of an 80 kg human body to cool it by 1°C? The source uses specific heat capacity of the human body ≈ 0.1 cal g−1 °C−1 and latent heat of vaporization of water at body temperature = 577 cal g−1.
Heat to be removed:
Q = msΔT = 80 × 1000 × 0.1 × 1 = 8000 cal
Heat removed by evaporation:
Q = mwLv = 577mw
577mw = 8000
mw = 13.86 g ≈ 0.0138 kg
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