1.Jack takes out a pack of sushi of mass 500g from a 2°C refrigerator, but he just leaves it on a table. Germs breed so quickly at 22°C that once the sushi reaches this temperature, it will be unsuitable for eating. If the specific heat capacity of the sushi is 3kJkg^-1°C^-1 and it gains heat at an average rate of 7W at room temperature of 25°C, within what time must Jack eat the sushi after taking it out the refrigerator?
2.‘Two rigid bodies A and B at different temperatures are brought into contact.’ Assume no energy is lost to the surroundings. Based on the above statement, two students make the following claims. With whom do you agree? Why?
Ivan: By the law of conservation of energy, the sum of their internal energy must be the same as before.
Joe: By the law of conservation of energy, the potential energy lost by the molecules of the hotter body must be equal to the potential energy gained by the molecules of the colder body.
3.The high specific heat capacity of water not only enables it to absorb excessive heat, but also to store or release energy with only small temperature changes. Name one example of this.
4. A metal block is first immersed in boiling water for some time. The block is then transferred to a cup of cold water. After a while the temperature of the water is measured.
The result is as follows:
Mass of the metal block:0.8kg
Mass of water in the cup:0.3kg
Initial temp. of the water in the cup:23°C
Final temp. of water in the cup:38°C
Find the specific heat capacity of the metal(in Jkg^-1K^-1)
(Given the specific heat capacity of water:4200 Jkg^-1K^-1)
- 六呎將軍Lv 71 decade agoFavorite Answer
(1) Amount of heat that the sushi has to gain is:
0.5 x (22 - 2) x 3000 = 30000 J
Thus time period within which the sushi must be eaten = 30000/7 = 4286 s = 1.2 hr
(2) Ivan is correct. Since there's no energy loss to the surroundings, total energy of the objects is conserved which include both kinetic and potential energy.
Joe is incorrect since the hotter object's molecules may lose both potential (when changing state) and kinetic (when temperature falls) energy. Also the cooler object's molecules may gain both potential and kinetic energy.
(3) Absorbing heat: A good coolant for machines of car
Keeping temperature: Sea temperature is cooler in daytime and warmer in night time.
(4) Suppose there's no heat loss to the surroundings:
Heat gain by water = 4200 x 0.3 x (38 - 23) = 18900 J
So heat loss by metal block = 18900 J
Temp. change of metal block = 100 - 38 = 62 K
Sp, heat capacity of metal = 18900/(62 x 0.8) = 381 J/kg KSource(s): Myself