A steam boiler is made of steel and weighs 900lb. The boiler contains 400lb of water. How many Btu are required to raise the temperature of each from 42fahrenheit to 212fahrenheit? specific heat of steel = 0.11
To find the amount of heat required to raise the temperature of the steel and water in the steam boiler, you can use the formula:
Q = mcΔT
Where:
Q = heat energy (in British Thermal Units or BTU)
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
Let's calculate the heat required for each component separately:
For the steel:
mass = 900 lb (given)
specific heat capacity of steel = 0.11 (given)
change in temperature (ΔT) = 212°F - 42°F = 170°F
Q_steel = mass_steel × specific heat_steel × ΔT
= 900 lb × 0.11 × 170°F
= 20,790 BTU
For the water:
mass = 400 lb (given)
specific heat capacity of water = 1 BTU/lb°F (standard value for water)
change in temperature (ΔT) = 212°F - 42°F = 170°F
Q_water = mass_water × specific heat_water × ΔT
= 400 lb × 1 BTU/lb°F × 170°F
= 68,000 BTU
Therefore, the total heat required to raise the temperature of both the steel and water in the steam boiler is:
Total heat = Q_steel + Q_water
= 20,790 BTU + 68,000 BTU
= 88,790 BTU
So, 88,790 BTU are required to raise the temperature of the steel and water in the steam boiler from 42°F to 212°F.