How much calories does it take to raise the temperature of 225 g of H2O from 25 °C to 100 °C. (Specific heat of water 4.182 J/(g* °C)
I would use specific heat water as 1 cal/g*C and not do the Joule to calories conversion.
q = cal = mass H2O x specific heat H2O x (Tfinal-Tinitial).
If you want to use 4.182 J/g*C, the do that and convert to cal by
1 cal = 4.182 J.
To calculate the amount of calories needed to raise the temperature of water, you can use the specific heat formula:
Q = m * c * ΔT
Where:
Q is the amount of heat energy needed (in calories),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/(g*°C)),
ΔT is the change in temperature.
In this case, you are given the mass (m) of 225 g of water, the specific heat capacity of water (c) as 4.182 J/(g*°C), and the change in temperature (ΔT) from 25 °C to 100 °C.
Let's plug the values into the formula:
Q = 225 g * 4.182 J/(g*°C) * (100 °C - 25 °C)
Q = 225 g * 4.182 J/(g*°C) * 75 °C
Now we can calculate the answer.
Q = 70439.5 J
To convert this value to calories, we divide by 4.184 (since 1 calorie is equal to 4.184 joules):
Q = 70439.5 J / 4.184
Q ≈ 16,853.5 calories
Therefore, it takes approximately 16,853.5 calories to raise the temperature of 225 g of water from 25 °C to 100 °C.