F3 Physics (ENG)

2010-10-21 2:30 am
1. 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℃into 800g of water at 20℃. After a few minutes, he heats the fish-ball-amd-water "mixture" over a stove. Take the specific heat capacity of the fish ball as 2400 J kg^-1 ℃^-1 and the specific heat capacity of the water as 4200 J kg^-1 ℃^-1.
(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℃, what is the power of the stove?

(b). He then puts 500g of noodles at 15℃into the water. It takes another 1 minutes to heat the "mixture" up to 90℃ using the same setting. What is the specific heat capacity of the noodles?

2. In a freezing winter, a hot water bag with 300g of water is placed in an open area. The temperature of the water inside the hot water bag is 80℃and the surrounding temperature is 12℃. After 15 minutes, the temperature of the water becomes 65℃. How much energy is released to the surroundings from the hot water?
(The specific heat capacity of water is 4200 J kg^-1 ℃^-1. Assume the specific heat capacity of the bag itself is negligible.)

PS 請列出步驟!!!

回答 (1)

2010-10-21 3:54 am
✔ 最佳答案
1.(a)(i) 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
2. Heat released by hot water= (300/1000) x 4200 x (80-65) J= 18900 J


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