To calibrate the calorimeter electrically, a constant voltage of 3.6 v is applied and a current of 2.6 a flows for a period of 350 seconds. if the temperature rises from 20.3°c to 29.1°c, what is the heat capacity of the calorimeter? a) 1.1 x 102 j/°c b) 1.6 x 102 j/°c c) 3.7 x 102 j/°c d) 1.2 x 101 j/°c e) none of the above

Answer : The correct option is, (c)

Explanation :

First we have to calculate the energy or heat.

Formula used :

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

where,

C = heat capacity of the calorimeter

= initial temperature =

= final temperature =

Now put all the given values in this formula, we get:

Therefore, the heat capacity of the calorimeter is,

real, smaller and upside-down.   so the answer should be b.

see earlier