A 46kg sample of water absorbs 4.00×102kJ of heat. If the water was initially at 29.7 ∘ C , what is its final temperature?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
To find the final temperature of the water, we need to use the specific heat formula:
q = mcΔT
Where:
q is the amount of heat absorbed (4.00×10^2 kJ)
m is the mass of the water (46 kg)
c is the specific heat capacity of water
ΔT is the change in temperature
The specific heat capacity of water is approximately 4.18 kJ/(kg⋅°C).
Rearranging the formula, we have:
ΔT = q / (mc)
Substituting the given values, we have:
ΔT = (4.00×10^2 kJ) / (46 kg × 4.18 kJ/(kg⋅°C))
Now let's calculate:
ΔT = (4.00×10^2) / (46 × 4.18)
ΔT ≈ 2.941 °C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 29.7 °C + 2.941 °C
Final temperature ≈ 32.641 °C
Therefore, the final temperature of the water is approximately 32.641 °C.