F3 Physics Question(2)

2010-11-07 2:51 am

1. A kettle heats up 450-g water from 10°C to 90°C in 1.5 minutes. What is the power of the kettle? (The specific heat capacity of water is 4200 J kg^-1 °C^-1.) (3 marks)

2. When 6750J of energy has been transferred to a 3-kg metal block, the temperature of the block rises by 5°C. What are the specific heat capacity and heat capacity of the block? (4 marks)

10. A student sets up the apparatus shown to measure the specific heat capacity of an aluminium block.

The student obtains the following results:
Mass of aluminium block m = 0.8 kg
Temperature change T = 19°C
Time taken t = 5.0 minutes
Heater current I = 4.2A
Heater voltage V = 12V

(a) Given power = voltage X current, show by calculation, that 15120J of electrical energy are supplied to the heater in 5.0 minutes. (2 marks)

(b) (i) Assuming all of the electrical energy is transferred to the aluminium block as heat energy, calculate the value of the specific heat capacity of aluminium obyained from this experiment. (2 marks)
(ii) The accepted value of the specific heat capacity of aluminium is
902 J kg^-1 °C^-1.
(1) Give a reason for the difference between your answer in (b)(i) snd this value. (1 mark)
(2) How could the experiment be improved to reduce this difference?(1 mark)

回答 (1)

天同

2010-11-07 8:23 pm

✔ 最佳答案

1. Heat absorbed by water = (450/1000) x 4200 x (90 - 10) J= 151 200 J
Power = 151200/(1.5 x 60) w = 1680 w
2. Let c be the specific heat capacity of the blockHeat absorbed by block = 3c x 5 Hence, 15c = 6750c = 450 J/kg-C
10. (a) Power = 12 x 4.2 w = 50.4 wEnergy delivered in 5 min = 50.4 x (5x60) J = 15 120 J(b) (i) Let c be the specific heat capacity of aluminiumHeat absorbed by the block = 0.8c x 19hence, 0.8c x 19 = 15120c = 995 J/kg.C
(ii) (1) Not all energy produced by the heater was absorbed by the aluminium block. Hence, the rise in temperature was lower than it should be. Thus leads to a higher experimental value of specific heat capacity.

(2) Since you did not give the diagram of the experiment, an answer cannot be provided. The general approach, of course, is to reduce the heat loss from the heater.

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