Question

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?

Answer 1

`372.3 J/^(\circ)C`

Step-by-step explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

`P=VI=(3.6)(2.6)=9.36 W`

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

`E=Pt=(9.36)(350)=3276 J`

Finally, the change in temperature of an object is related to the energy supplied by

`E=C\Delta T`

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

`\Delta T =29.1^(\circ)-20.3^(\circ)=8.8^(\circ)C` is the change in temperature

Solving for C, we find:

`C=(E)/(\Delta T)=(3276)/(8.8)=372.3 J/^(\circ)C`