Answer:
83.13 degree Celsius folks!
Step-by-step explanation:
Q = msdT
Here is how you do it fella's! First off Let us know abt the above terms shall we?
Where:
Q is the heat energy absorbed by the water,
m is the mass of the water,
s is the specific heat capacity of water,
dt is the change in temperature.
Given:
Q = 3.6 x 10^5 J
m = 2 kg
s= 4,186 J/kg°C (specific heat capacity of water)
Initial temperature, T_initial = 40°C
Rearranging the formula, we have:
dt = Q / (ms)
Substituting the given values:
dt = (3.6 x 10^5 J) / (2 kg * 4,186 J/kg°C)
Calculating:
dt ≈ 43.13°C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = T_initial + dT(above formed temp)
Final temperature = 40°C + 43.13°C
Final temperature = 83.13°C
Therefore, the correct final temperature of the water is approximately 83.13.