Energy in the amount of 479 J is added to a
89.0 g sample of water at a temperature of
8.00◦C. What will be the final temperature of
the water?
Answer in units of ◦C.
Q=mcΔT
Q=mc(Tf-Ti)
(Tf-Ti)=Q/mc
Tf=(Q/mc)+Ti
Tf=(479/(89.0)(4.184))+8.00
To find the final temperature of water after adding energy, we can use the specific heat equation. The equation is as follows:
q = m * c * ΔT
Where:
q is the heat energy absorbed or lost by the substance
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we are given:
q = 479 J (the heat energy added to the water)
m = 89.0 g (the mass of the water sample)
ΔT = final temperature - initial temperature
First, we need to calculate the heat energy absorbed by the water.
q = m * c * ΔT
479 J = 89.0 g * c * ΔT
Next, we will use the specific heat capacity of water, which is approximately 4.18 J/g°C. Substituting this value into the equation:
479 J = 89.0 g * 4.18 J/g°C * ΔT
Now, we can solve for ΔT (the change in temperature).
479 J = 372.02 g°C * ΔT
ΔT = 479 J / (372.02 g°C) ≈ 1.29°C
Finally, to find the final temperature, we add the change in temperature to the initial temperature.
Final temperature = Initial temperature + ΔT
Final temperature = 8.00°C + 1.29°C ≈ 9.29°C
Therefore, the final temperature of the water will be approximately 9.29°C.