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Gnoma [55]
2 years ago
6

PLEASE HELP I WILL GIVE BRALINEST

Physics
1 answer:
Greeley [361]2 years ago
6 0
Weight: it is the gravitational pull applied on a body by the earth’s surface
Mass: it is the amount of matter present
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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 3
iren [92.7K]

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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