3 years ago

How much heat is gained by 1.0 gram of iron when its heated 15 degrees celsius

1 answer:

3 0

Q = mass water x specific heat water x delta T.
<span>714,000 = mass water x specific heat water x 30.
Substitute specific heat water and solve for mass water.</span>

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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

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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

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Solving for C, we find:

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

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3 years ago

A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco

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S=(0x4)+(0.5x4.81x4x4)
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