How many Btu of heat must be added to 900lbs of steel to raise its temperature from 20.0°F to 500°F?
Specific heat of steel .115Btu/lb°F
Q= c•m •Δt =0.115•900•480 =49680 Btu
To find out how many Btu of heat must be added to 900 lbs of steel, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat energy (in Btu)
m = mass of the substance (in lbs)
c = specific heat of the substance (in Btu/lb°F)
ΔT = change in temperature (in °F)
Given:
m = 900 lbs
c = 0.115 Btu/lb°F
ΔT = 500°F - 20°F = 480°F
We can substitute the given values into the formula:
Q = 900 * 0.115 * 480
Calculating this, we have:
Q = 49,680 Btu
Therefore, 49,680 Btu of heat must be added to 900 lbs of steel to raise its temperature from 20.0°F to 500°F.