If 335g water at 65.5 degC loses 9750 J of heat, what is the final temperature of the water?

q = mass x specific heat water x (Tfinal-Tinitial). Solve for Tfinal

To determine the final temperature of the water, you can use the formula for heat transfer:

Q = mcΔT

where:
Q = heat transferred (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (in °C)

First, let's rearrange the formula to solve for the change in temperature:

ΔT = Q / mc

Now we can plug in the given values:
Q = 9750 J
m = 335 g
c = 4.18 J/g°C (specific heat capacity of water)

Substituting these values into the formula, we have:

ΔT = 9750 J / (335 g * 4.18 J/g°C)

Calculating this expression, we get:

ΔT ≈ 7.41°C

Finally, to find the final temperature, we add the change in temperature to the initial temperature of 65.5°C:

Final temperature = Initial temperature + ΔT
Final temperature = 65.5°C + 7.41°C

Therefore, the final temperature of the water is approximately 72.91°C.