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How much heat is needed to raise the temperature of 4.68 kg of lead from 25°C to 62°C

1 Answer

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The specific heat capacity of lead is approximately 0.128 J/g°C. To calculate the heat needed, you can use the formula:

Q = m * c * ΔT

Where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)

First, convert the mass to grams: 4.68 kg = 4680 g

Now, plug the values into the formula:
Q = 4680 g * 0.128 J/g°C * (62°C - 25°C)

Calculate the result to find the heat energy needed to raise the lead's temperature:
Q = 4680 g * 0.128 J/g°C * 37°C = 22677.12 joules

So, approximately 22677.12 joules of heat energy is needed to raise the temperature of 4.68 kg of lead from 25°C to 62°C.
User Gaspar Teixeira
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