A 5.0×101 kg sample of water absorbs 324 kJ of heat. If the water was initially at 22.2 ∘C, what is its final temperature?
324kj=50kg(Cwater)(Tf-22.2)
look up Cwater, check units (you want it in kj/kg), then solve for Tf
To calculate the final temperature of the water, we need to use the formula for heat exchange:
Q = mcΔT
where:
Q is the heat absorbed or released (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
In this case, we are given:
Q = 324 kJ (which we need to convert to joules),
m = 5.0 × 10^1 kg,
c is the specific heat capacity of water, which is about 4.18 J/g·°C.
First, we need to convert kilojoules (kJ) to joules (J). Since 1 kJ = 1000 J, we multiply the given heat, 324 kJ, by 1000:
Q = 324 kJ × 1000 J/kJ = 324000 J
Next, we plug in the values into the formula and solve for ΔT:
324000 J = (5.0 × 10^1 kg) × (4.18 J/g·°C) × ΔT
Now, let's simplify the equation:
324000 J = (5.0 × 10^1 kg) × (4.18 J/g·°C) × ΔT
324000 J = (5.0 × 10^1 kg) × (4.18 J/g·°C) × ΔT
Now, let's cancel out the units to make it easier to solve:
324000 J = (5.0 × 10^1 kg) × (4.18 J/g·°C) × ΔT
324000 J = (5.0 × 10^4 g) × (4.18 J/g·°C) × ΔT
Next, let's solve for ΔT:
ΔT = 324000 J / [(5.0 × 10^4 g) × (4.18 J/g·°C)]
ΔT ≈ 1.55 °C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = 22.2 °C + 1.55 °C = 23.75 °C
Therefore, the final temperature of the water is approximately 23.75 °C.