What is final temperature of a 88 gram sample symbol of water at 295 k that absorbs 1250 Joules of heat ?
1250 J = mass H2O x specific heat x (Tfinal-Tinitial)
Solve for Tf.
Thank you ! Do I have to K to C right ?
To find the final temperature of the water sample, we can use the equation:
Q = mcΔT
Where:
Q = heat energy added (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
In this case, we know:
Q = 1250 J
m = 88 g
c = specific heat capacity of water (4.18 J/g°C)
We are given the initial temperature (295 K), which we need to convert to °C:
T(°C) = T(K) - 273.15
T(°C) = 295 - 273.15
T(°C) = 21.85°C
Now, let's plug in the values into the equation:
1250 = 88 * 4.18 * ΔT
Simplifying the equation:
1250 = 367.84 * ΔT
Dividing both sides by 367.84:
ΔT = 1250 / 367.84
ΔT ≈ 3.4°C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 21.85°C + 3.4°C
Final temperature ≈ 25.25°C
Therefore, the final temperature of the 88 gram sample of water, after absorbing 1250 Joules of heat, is approximately 25.25°C.