F.3 Physics 高手請進 Heat

2012-12-09 12:51 am

1.An iron sphere of mass 1.2kg is cooled by immersing it in a tank of water.
The mass of water is 3 kg and its initial temperature is 126度.After a certain
time, both the iron sphere and the water are at 15度.
(Specific heat capacity of iron=480;
Specific heat capacity of water=4200)

2.(A) If we submerge a 2-kg steel block at 90度 into 3kg of water at10度,what will be the temperature of the water?
(Specific heat capacity of steel=450;
Specific heat capacity of water=4200)
(B) Suggest one advantage of using water as a coolant in engines.

3.Baba is going to make a bowl of fish ball noodles. The mass of each fish ball
is 80g.
(A)Baba puts two fish balls at 2度 into 800g of water at 20度.After a few minutes, he heats the fish-ball-and-water ‘mixture’ over a stove. Take the specific capacity of the fish ball as 2400 and the specific heat capacity of water as 4200.
(i) What is the temperature of the mixture before it is heated?
(ii) If it takes 5 minutes to heat the ‘mixture’ up to 90度, what is the power of the stove?
(B)He then puts 500g of noodles at 15度into the water. It takes another 1 minutes to heat the ‘mixture’ up to 90度 using the same setting. What is the specific heat capacity of the noodles?

更新1:

1.更改: An iron sphere of mass 1.2kg is cooled by immersing it in a tank of water. The mass of water is 3 kg and its initial temperature is 12度.After a certain time, both the iron sphere and the water are at 15度. (Specific heat capacity of iron=480; Specific heat capacity of water=4200)

回答 (2)

Mimi

2012-12-14 6:54 am

✔ 最佳答案

1.Let the initial temperature of iron sphere be y
2(480)(y-15)= 3(4200)(15-12)
y=54.4(corr to 3 sig fig)

2.Let the final temperature of water be y
2(450)(90-y)=3(4200)(y-10)
(90-y)/(y-10)=14
90-y=14y-140
230=15y
y=15.3( corr to 3 sig fig)

Water has a high specific heat capacity, it can bring away much heat energy from engines without its own temperature being raised too much.

3. Let T be the temperature of the mixture.(2 x 80/1000) x 2400 x (T-2) = (800/1000) x 4200 x (20-T)
solve for T gives T = 18.15'C

(ii) Heat absorbed in 5 minutes
= [(2x80/1000) x 2400 x (90-18.15) + (800/1000) x 4200 x (90-18.15)] J = 269,000 J
Power of stove = 269000/(5x60) watts = 897 watts

(b) Heat given out by the stove in 1 minute = 897 x 60 J = 53820 JHence, specific heat capacity of noodels
= 53820/[(0.5 x (90-15)] J/kg-C = 1435 J/kg-C

2012-12-13 22:54:30 補充：
http://hk.knowledge.yahoo.com/question/question?qid=7009100601225

參考: http://hk.knowledge.yahoo.com/question/question?qid=7010102000815

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天同

2012-12-09 1:39 am

1. How come the temperature of water is 126 deg.C? This is already above the water boiling point.

2. Let T be the final temperature of water.
By the heat balance equation, heat lost by steel block = heat gained by water
2 x 450 x (90-T) = 3 x 4200 x (T-10)
solve for T gives T = 15.33'C

3.(a)(i) Use the heat balance equation, let T be the temperature of the mixture.
0.8 x 4200 x (20 - T) = 2x0.08 x 2400 x (T - 2)
solve for T gives T = 18.15'C

(ii) Heat aborbed by the mixture
= (0.8 x 4200 + 2x0.08 x 2400) x (90 - 18.15) J = 2.69 x 10^5 J
Power of stove = 2.69 x 10^5/(5x60) watts = 897 watts

(b) Let s be the specific heat capacity of the noodle,
0.5s x (90 - 15) = 897 x 60
s = 1435 J/kg.C

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