How much heat is required to warm 250.4 g of water from 22.6 C to 75.8 C given that the specific heat capacity of water is 4.184 J/gx degrees C?
q = mass x specific heat x delta T.
55736.25
55736.25 i think not sure
To calculate the heat required to warm a substance like water, you can use the formula:
q = m * c * ΔT
where:
q is the heat energy (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
In this case, we have:
m = 250.4 g
c = 4.184 J/g°C
ΔT = (75.8°C - 22.6°C)
First, calculate the change in temperature:
ΔT = 75.8°C - 22.6°C = 53.2°C
Substitute the values into the formula and solve for q:
q = 250.4 g * 4.184 J/g°C * 53.2°C
q ≈ 56,036.1472 J
So, approximately 56,036.1472 Joules of heat are required to warm 250.4 grams of water from 22.6°C to 75.8°C.