✔ 最佳答案
4. Let m1 and m2 be the masses of copper and zinc respectively.
Consider one degree C rise in temperature,
heat absorbed by copper = 380(m1)
heat absorbed by zinc = 388(m2)
heat absorbed by brass = 385.8(m1 + m2)
Hence, 380(m1) + 388(m2) = 385.8(m1 + m2)
i.e. 380(m1/m2) + 388 = 385.8(m1/m2) + 385.8
(m1/m2) = 0.38
5. Let Q be the heat supplied and m be the mass of water vapourized.
L be the latent heat of capourization of water.
Hence, Q = mL
i.e. L = Q/m
Statement 1 is correct.
When Q is overestimated, L(experiment) is higher than L(true).
Statement 2 is correct.
The measured value of m is lower than the true value. Hence, L(experiment) > L(true).
Statement 3 is wrong.
The measured value of m is higher than the true value.
Hence, L(experiment) < L(true)
The answer is option C.
2014-08-04 07:40:48 補充：
Your suppl questions:
380(m1) + 388(m2) = 385.8(m1 + m2)
Divide each term by m2
380(m1/m2) + 388(m2/m2) = 385.8(m1/m2) + 385.8(m2/m2)
380(m1/m2) + 388 = 385.8(m1/m2) + 385.8
2014-08-04 07:45:03 補充：
Q:想問番(2) option 果度,water condensing on the cooler part of the beaker……d水會唔會流番落去個beaker度?我只係想知甘解得唔得。
A: Yes. The condensing water will go back to the beaker. Thus the measured mass of water loss due to evapouration is less than it should be. This leads to an overestimation of the latent heat.
