physic 問題 快進!!!!!!!(3) help me

HI
BY C=Q/delta(個三角形)T
C=0.5x4200x58/(58-26)
C=121800/32
C=3806.25
=3806j c^-1(corect to nearest integer)

The question ask for heat capacity ,but your answer is in terms of specific heat capacity.There must be something wrong.

Let H be the heat capacity of the thermometer
hence, heat gained by thermometer = H.(58-26) = 32H

Heat lost by the water = 0.5 x 4200 x (60 - 58) J = 4200 J
Assume no heat lost to the surroundings,
32H = 4200
i.e. H = 4200/32 J/C = 131 J/C

[Your given answer is probably wrong as it is a value for the specific heat capacity, but you are asked to find the "heat capacity" of the termometer].

2011-06-11 14:55:41 補充：
I still don't think your answer of 3806 J/C is right. The value seems too high and is not realistic. A practical thermometer should have a low heat capacity so as to give an accurate temperature reading.

The mass of the thermometer should also be provided.