A 5.0×101kg sample of water absorbs 3.10×102kJ of heat. If the water was initially at 21.3 ∘ C, what is its final temperature?
To find the final temperature of water after absorbing a certain amount of heat, we can use the formula:
q = mcΔT
Where:
q is the heat absorbed by the water (in Joules)
m is the mass of the water (in kg)
c is the specific heat capacity of water (4.18 J/g°C or 4186 J/kg°C)
ΔT is the change in temperature (final temperature - initial temperature) in Celsius.
First, we need to convert the mass of water from kg to grams:
5.0 × 10^1 kg = 5.0 × 10^4 g
Now, we can substitute the given values into the formula and solve for ΔT:
3.10 × 10^2 kJ = (5.0 × 10^4 g) × (4.18 J/g°C) × ΔT
Next, we need to convert the given quantity of heat from kilojoules to joules by multiplying by 1000:
3.10 × 10^2 kJ = 3.10 × 10^5 J
Substituting the values and solving for ΔT:
3.10 × 10^5 J = (5.0 × 10^4 g) × (4.18 J/g°C) × ΔT
Now, let's solve for ΔT:
ΔT = (3.10 × 10^5 J) / [(5.0 × 10^4 g) × (4.18 J/g°C)]
ΔT ≈ 14.81°C
To find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 21.3°C + 14.81°C
Final temperature ≈ 36.11°C
Therefore, the final temperature of the water is approximately 36.11°C after absorbing 3.10 × 10^2 kJ of heat.
q = 310,000 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute and solve for Tfinal.