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A 28 g coin absorbs 658 J of heat while it increases in temperature from 25 to 125 °C what is the specific heat of the metal in the coin?

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User Jen Grant
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Answer: We can use the formula for heat (Q) gained or lost by an object undergoing a temperature change:

Q = mcΔT

where Q is the heat absorbed or released, m is the mass of the object, c is the specific heat of the object, and ΔT is the change in temperature.

We are given that a 28 g coin absorbs 658 J of heat while it increases in temperature from 25 to 125 °C. We can plug these values into the formula and solve for c:

658 J = (0.028 kg) * c * (125 °C - 25 °C)

Simplifying, we get:

658 J = 0.028 kg * c * 100 °C

Dividing both sides by (0.028 kg * 100 °C), we get:

c = 658 J / (0.028 kg * 100 °C)

c ≈ 234.6 J/(kg·°C)

Therefore, the specific heat of the metal in the coin is approximately 234.6 J/(kg·°C).

Explanation:

User Stuart Robertson
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