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.

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?

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Thu Thủy 5 years 2021-09-04T22:54:55+00:00 1 Answers 112 views 0

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  1. Maris
    0
    2021-09-04T22:56:30+00:00
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    Answer:

    372.3 J/^{\circ}C

    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=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

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