有physics 問題唔識

2009-10-02 4:49 am
1.A bowl of soup of mass 200g cools down from 85°C to 50°C in 28 mins. It is given that the specific heat capacity of the soup is 400J kg^(-1)°C^(-1).
a)Find the energy loss to the surroundings.

b)Find the average rate of energy loss by the soup.


2.An air conditioner cools air by removing energy from the room to the outdoors. Its cooling capacity is the amount of energy drawn out per second. Take the specific heat capacity of air to be 1000J kg^(-1)°C^(-1).
a)An air conditioner in an office has a cooling capacity of 3.5kW. The mass and the initial temp. of the air inside are310kg and 32°C respectively. Suppose the room temp. is lowered to 22°C, how long is the air conditioner switched on? Assume there is no heat flow other than that cause by the air conditioner.

b)A commercial complex has 4000 such offices.Each office installs an air conditioner which is mentioned in (a). If all the rooms are to be cooled from 32°C to 22°C at the same time, how much energy will be drawn out? If this energy is used to heat up a swimming pool of water of mass 3*10^5 kg, what will be the temp. rise? Take the specific heat capacity of the water to be 4200J kg^(-1)°C^(-1).

回答 (1)

2009-10-02 5:05 am
✔ 最佳答案
1.a. Energy loss to the surroundings

= mcT

= (0.2)(400)(85 - 50)

= 2800 J

b. Average rate of energy loss

= Energy lost / Time

= 2800 / (28 X 60)

= 1.67 Js-1

2.a. Energy loss by the air

= mcT

= (310)(1000)(32 - 22)

= 3 100 000 J

Time required = Energy / Power

= 3 100 000 / 3500

= 886 s

= 14.8 mins

b. Total amount of energy = 3 100 000 X 4000 = 1.24 X 1010 J

E = mcT

1.24 X 1010 = (3 X 105)(4200)T

Temperature rise, T = 9.84 *C

參考: Physics king


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