Physics heat 問題

Basically I agree with the previous answer given by 天同. However as that asnwer is not a choice given, and want to answer one possiblilty that I can think of.

My answer will be A. (1) only.

We all know that water is being loss to the surrounding by evaporation, and evaporation is related to the surface area of the tank and the temperture difference between water and the surrounding. As the tank is the same tank, and there is no heat loss to the surrounding, we can assume the rate of water loss to the surrounding is also constant.

What if the rate of water loss to the surrounding via evaporation is the same as the rate of water filling the tank? With a large surface area tank and a low surrounding temperature, this is possible. In such case, the volume of water in the tank will be constant. Then we will have two cases. (a) the heater heat up the water faster than the cooling by addition of water --> hence graph (2) only. (b) the heater heat up the water at a rate same as the addtion of water cooling the water --> hence graph (1) only.

As (2) only is not a choice given. THe answer can only be A. (1) only.


P.S. But I too think 天同 is correct. It is probably an error in the exercise.

2013-05-22 03:08:17 補充：
Referring to the question, actually it was already written in my previous answer, but I can write it again.

(1) The heat was used up by the evaporation of water.
(2) The volume of water is contant due to evaporation.

2013-05-22 03:08:36 補充：
It was a possible but highly unlikely case. But if you learn physics, you know physics is about what won't actually happened... it always needs assumption, for example assume no heat lost, assume standart temperature and pressume, assume no fiction... etc.

2013-05-22 03:08:40 補充：
It need all kind of assumption, which mean the case is never real. I was just trying to point out the only possible case for this question with possible assumptions in order to fulfill the choices of answers. Thank you for your question. =)

2013-05-23 17:03:24 補充：
標準答案是 A ，我不是解釋到為何是 A 嗎?

I don't think any one of the curve is correct.

Suppose the power of the heater = P
The inital mass of water in the tank = M
The rate of water filling in the tank = k
Hence, the mass of water in the tank at time t = kt + M

Since there is no heat loss to the surroundings, all heat produced by the heater is being absorbed by the water. Using the heat balance equation,

P (Δt) = (kt + M).c.(ΔT)
where Δt is an infinitesimal time interval, and ΔT is the corresponding rise in temperature T, and c is the specific heat capacity of water

Thus, ΔT/Δt = P/[(kt + M).c]
But (ΔT/Δt) is the slope of the curve of temperature T against time t

Therefore, it can be seen that the slope of the curve decreases with time t. None of the 3 curves shown demonstrates such feature.





2013-05-20 10:36:05 補充：
Thanks for Kwan941332 who brought out some ideas. But I don't think both (1) and (2) could be correct.

2013-05-20 10:41:38 補充：
For (1), the question clearly says that the water added in is at 25'C, which I presume is the same temperature as water originally in the tank. The newly added water thus DOES NOT need to be heated up. Then, where does the heat produced by the heater go?

2013-05-20 10:42:08 補充：
(cont'd)..A constant temperature of water, as shown in graph (1), does not require any heat.

2013-05-20 10:45:03 補充：
For graph (2), a linear rise of temperature with time is only possible when the mass of water in the tank remains constant. But this is not the case in this problem, as the mass of water is increasing with time.

2013-05-20 10:49:56 補充：
(cont'd)... As such, we would expect the temperature rise gradually slows down with the increasing of water mass. This also precludes the correctness of graph (3).