Lesson 38 39 answers – Specific Heat Capacity
( /33)……..%…….
ALL
1 What unit is used for all types of energy?
………Joule…………………………………………………………………………… (1)
2 State the equation for the energy needed to raise the temperature of a particular material of known mass.
…………Energy = mass x specific heat capacity x change in temperature……………
………………………………………………………………………………………… (1)
3 Use the following data to calculate the amounts of energy needed to change the temperature of the questions below.
Material / Specific heat capacity (J/kg ˚C)Copper / 380
Water / 4200
Aluminium / 880
a) 2kg of water by 5˚C
…………2 x 380 x 5 = 3800 J…………………………………………………………
………………………………………………………………………………………… (2)
b) 500g of water by 4˚C
………0.5 x 420 x 4 = 8400 J……………………………………………………………
………………………………………………………………………………………… (2)
c) 100g of aluminium from 20˚C to 30˚C
………0.1 x 880 x 10 = 880 J…………………………………………………
………………………………………………………………………………………… (2)
d) 200 g of copper from 60˚C to 10˚C
……….0.2 x 380 x –50 = -3800 J
(negative because it is the energy given out by the copper)………………………………
………………………………………………………………………………………… (3)
4 A 2kg block of iron is given 10kJ of energy and its temperature rises by 10˚C. What is the specific heat capacity of iron?
……… Energy = mass x specific heat capacity x change in temperature
………So, specific heat capacity = Energy / (mass x change in temperature)
…………… specific heat capacity = 10000 / (2 x 10)
…………… specific heat capacity = 500 J / kg˚C…………………………………… (4)
5 Some cooks make toffee. Essentially, this is a process of boiling down a sugar solution to concentrate it and then allowing the liquid to cool until it sets. Small children are usually warned not to touch the cooling toffee for a very long time – much longer than the cooling for the same volume of pure water in the same vessel. Why do you think that the cooling period so long?
…… There are a number of factors that determine the time it takes the toffee to cool sufficiently to eat:
The boiling point of the sugar solution is higher than that of water so the toffee is cooling from a higher temperature.
The sugar solution has a higher specific heat capacity than pure water. So for an equivalent temperature drop, more energy has to be lost.
[Perhaps most significant is the amount of energy that has to be lost for the liquid toffee mixture to solidify, cooling and then changing from a liquid to a solid takes a long time. This involves the idea of latent heat capacity, rather than specific heat capacity] (2)
6 Show that the energy required to heat the air in your physics laboratory from a chilly 10 °C to a more comfortable 20 °C is about 3 000 000 J if it has the following dimensions: 3 m ´ 10 m ´ 10 m.
Assume the laboratory is 3 m ´ 10 m ´ 10 m, this leads to a volume of 300 m2. Assume further that the heating is just a question of warming up the air. The density of air is approximately 1 kg m–3 so energy = 300 kg ´ 1000 J kg–1°C–1 ´ 10°C, i.e. 3 MJ.
A reasonable heater might deliver this in 1000 s (about 20 minutes). Most people would guess that the heating time would be much longer. This estimate ignores heating the contents of the room, etc. Our perception of temperature is affected both by the humidity of the air and by the cooling effect of any draughts. It would take much longer in reality. (4)
7 This question is about the operation of an electrically operated shower.
a) The water moves at constant speed through a pipe of cross section 7.5 x 10-5m2 to a showerhead. See the diagram. The maximum mass of water which flows per second is 0.090kgs-1.
i) Show that the maximum speed of the water in the pipe is 1.2ms-1.
Density of water = 1000kg m-3
Density = m/V
Volume flowing in 1 second = 7.5 x 10-5m2 x 1.2ms-1
Therefore 1000 x 0.00009 = mass/second = 0.090kg/s QED
(2)
ii) The total cross-sectional area of the holes in the head is half that of the pipe. Calculate the maximum speed of the water as it leaves the shower head.
Same volume must flow therefore speed must be double
Speed = ……2.4………… ms-1 (1)
iii) Calculate the magnitude of the force on the shower-head.
F = m (v-u)/t
F = 0.09 (2.4-1.2)
Force = ……0.11……….. N (3)
b) The water enters the heater at the temperature of 15˚C. At the maximum flow rate of 0.090kgs-3, the water leaves the shower head at a temperature of 27˚C.
i) Calculate the rate at which energy is transferred to the water. Give a suitable unit for your answer.
Specific heat capacity of water = 4200Jkg-1K-1
ΔQ=mcΔθ
= 0.09 x 4200 x (27-15)
Rate of energy transfer = ……4536………… unit ……W (Js-1)……….. (4)
ii) Suggest a reason that the power of the heater must be greater than your answer to (b)(i)
……Energy (heat ) loss to surroundings…………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
iii) Calculate the maximum possible temperature of the water at the showerhead when the flow rate is half of the maximum.
4536 / 0.045 x 4200 = 24
If start temp still 15
So Δθ = 15 + 24
Temperature = ………39………….˚C (1)