If 100.0 J are added to 20.0 grams of water at 30.0 degrees Celsius. What will be the final temperature?
q = mass x specific heat x (Tfinal-Tinitial)
To find the final temperature, we can use the principle of heat transfer, which states that the heat gained by one substance is equal to the heat lost by another substance.
In this case, the water gains heat and the energy gained is given as 100.0 J. We can solve for the final temperature by using the formula for heat gained or lost:
Q = m * c * ΔT
Where:
Q = heat gained or lost (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance
ΔT = change in temperature
We know:
Initial temperature (Ti) = 30.0°C
Mass (m) = 20.0 grams
Energy gained (Q) = 100.0 J
We need to find the final temperature (Tf).
We can rearrange the formula to solve for ΔT:
ΔT = Q / (m * c)
Since we need to find ΔT, we can calculate it using the given values:
ΔT = 100.0 J / (20.0 g * c)
The specific heat capacity of water is approximately 4.18 J/g°C.
Plug in the known values to find ΔT:
ΔT = 100.0 J / (20.0 g * 4.18 J/g°C)
Simplifying the expression, we get:
ΔT ≈ 1.20 °C
Finally, to find the final temperature (Tf), add the change in temperature (ΔT) to the initial temperature (Ti):
Tf = Ti + ΔT = 30.0°C + 1.20°C
Therefore, the final temperature will be approximately 31.20°C.