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,

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

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