If 311 g water at 65.5°C loses 9760 J of heat, what is the final temperature of the water?
-q = mass water x specific heat water x (Tfinal-Tinitial)
To find the final temperature of the water, we can use the equation:
q = mcΔT
Where:
q = heat gained or lost by the substance (in this case, water) (in J)
m = mass of the substance (in this case, water) (in g)
c = specific heat capacity of the substance (in this case, water) (in J/g°C)
ΔT = change in temperature (in this case, the final temperature minus the initial temperature) (in °C)
Let's break down the information given:
m = 311 g (mass of water)
ΔT = final temperature - initial temperature
We are given:
q = -9760 J (negative sign indicates heat lost by the water)
initial temperature = 65.5°C
First, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Now, let's rearrange the equation to solve for ΔT:
q = mcΔT
ΔT = q / mc
Substituting the given values:
ΔT = -9760 J / (311 g * 4.18 J/g°C)
Calculating:
ΔT = -9760 J / 1299.98 J/°C
ΔT ≈ -7.51°C
To find the final temperature, we need to subtract ΔT from the initial temperature:
Final temperature = 65.5°C - 7.51°C
Calculating:
Final temperature ≈ 57.99°C
Therefore, the final temperature of the water is approximately 57.99°C.