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How many BTUs of heat would be required to raise the temperature of 1/2 pound of lead from 65°F to its melting point of 621.5°F, given that the specific heat capacity of lead is 0.0305 BTU/°F·lb?

User Aolde
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1 Answer

3 votes

Final answer:

To raise the temperature of half a pound of lead from 65°F to its melting point at 621.5°F, 8.487775 BTUs of heat are needed.

Step-by-step explanation:

To calculate the number of BTUs of heat required to raise the temperature of 0.5 pounds of lead from 65°F to its melting point of 621.5°F, we can use the formula Q = mcΔT, where Q is the heat in BTUs, m is the mass in pounds, c is the specific heat capacity in BTU/°F·lb, and ΔT is the change in temperature in °F.

First, we need to calculate the temperature change:

ΔT = 621.5°F - 65°F = 556.5°F

Then, we plug our values into the formula:

Q = (0.5 lb) × (0.0305 BTU/°F·lb) × (556.5°F)

After multiplying the values, the answer is:

Q = 0.5 × 0.0305 × 556.5

Q = 8.487775 BTUs

Therefore, 8.487775 BTUs of heat are required to raise the temperature of 0.5 pounds of lead to its melting point.

User Lou Grossi
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