A 5.0×101 kg sample of water absorbs 333 kJ of heat.
If the water was initially at 23.1 ∘C, what is its final temperature?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
mass is 50 kg. Chane to g.
q = 333 kJ. Change to J.
Substitute into the equatiion and solve for Tfinal.
Specific heat H2O should be in units of J/g*C which is 4.184 J/g*C
Post your work if you get stuck.
To find the final temperature of the water, we need to use the equation:
Q = mcΔT
Where:
Q = heat absorbed or released by the substance (in joules)
m = mass of the substance (in kg)
c = specific heat capacity of the substance (in J/kg°C)
ΔT = change in temperature (in °C)
In this case, we're given:
Q = 333 kJ = 333,000 J
m = 5.0 × 10^1 kg
c = specific heat capacity of water = 4.18 kJ/kg°C = 4,180 J/kg°C
ΔT = final temperature - initial temperature
We want to find the final temperature, so we rearrange the equation:
ΔT = Q / (mc)
Now we can substitute the given values:
ΔT = 333,000 J / (5.0 × 10^1 kg * 4,180 J/kg°C)
Calculating the right side of the equation yields:
ΔT ≈ 15.9 °C
Finally, we add the change in temperature to the initial temperature to find the final temperature:
Final temperature = Initial temperature + ΔT
Final temperature ≈ 23.1 °C + 15.9 °C
Final temperature ≈ 39.0 °C
Therefore, the final temperature of the water is approximately 39.0 °C.