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The specific heat of iron is 0.45 J/g C. How much heat is transferred when a 24.7 kg iron ingot is cooled from 880 C to 100 C?

User Dillon
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Answer:

The amount of heat transferred when a substance is cooled can be calculated using the formula:

Q = m * c * ΔT

Where:

Q is the heat transferred (in joules),

m is the mass of the substance (in grams),

c is the specific heat of the substance (in J/g°C), and

ΔT is the change in temperature (in °C).

In this case, we have an iron ingot with a mass of 24.7 kg, which is equivalent to 24,700 grams. The specific heat of iron is given as 0.45 J/g°C. The change in temperature is the difference between the initial temperature of 880°C and the final temperature of 100°C.

ΔT = 100°C - 880°C

ΔT = -780°C

Since the temperature has decreased, the change in temperature is negative.

Now, let's calculate the heat transferred:

Q = (24,700 g) * (0.45 J/g°C) * (-780°C)

Multiplying these values, we get:

Q = -8,297,800 J

Therefore, the amount of heat transferred when the 24.7 kg iron ingot is cooled from 880°C to 100°C is -8,297,800 Joules (or you can also write it as -8.3 MJ, which stands for megajoules).

It's worth noting that the negative sign indicates that heat has been lost from the iron ingot as it cooled.

User Dpigera
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