What is the mass of water that has its temperature raised from 27°C to 32°C when it is heated by 5.0 x 106 J of heat energy?
To find the mass of water, we need to use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the heat energy absorbed or released by the substance (in joules),
m is the mass of the substance (in grams or kilograms),
c is the specific heat capacity of the substance (in J/g°C or J/kg°C),
ΔT is the change in temperature (in °C).
In this case, we know the heat energy (Q) is 5.0 x 10^6 J, the initial temperature (T1) is 27°C, and the final temperature (T2) is 32°C. We need to find the mass (m) of water.
We also need the specific heat capacity of water, which is approximately 4.18 J/g°C or 4186 J/kg°C.
First, let's calculate the change in temperature:
ΔT = T2 - T1
= 32°C - 27°C
= 5°C
Now we can rearrange the formula to solve for mass (m):
m = Q / (c * ΔT)
Substituting the values we have:
m = 5.0 x 10^6 J / (4.18 J/g°C * 5°C)
m ≈ 238,993 g
Therefore, the mass of water that has its temperature raised from 27°C to 32°C when heated by 5.0 x 10^6 J of heat energy is approximately 238,993 grams.