What is the temperature change in celsius when 29.5L of water absorbs 2.3 kwh of heat?
1 watt = 1 J/second
2.3 kwh = 2.3E3 x (60 s/min) x (60 min/hr) = ?? J/
??J = mass H2O x specific heat H2o x (Tfinal-Tinitial)
I assume the water has density of 1.00 g/mL wo 29.5 L will have a mass of 29.5 x 1000 to convert to cc = 29500 grams.
To determine the temperature change in Celsius when a certain amount of heat is absorbed by water, we can use the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
First, we need to convert the amount of heat absorbed from kilowatt-hours (kWh) to joules (J).
1 kilowatt-hour (kWh) = 3,600,000 joules (J)
Therefore, 2.3 kWh = 2.3 × 3,600,000 J = 8,280,000 J of heat.
Next, we need to calculate the mass of water. The density of water is about 1 gram per milliliter (g/mL) or 1000 kilograms per cubic meter (kg/m³).
29.5 liters (L) = 29.5 × 1000 mL = 29,500 grams (g) = 29.5 kilograms (kg)
Now we have the heat energy (8,280,000 J) and the mass of water (29.5 kg).
Using the formula:
Q = m × c × ΔT,
where Q is the heat energy in joules, m is the mass of the substance in kilograms, c is the specific heat capacity of the substance, and ΔT is the temperature change in Celsius.
We can rearrange the formula to solve for ΔT:
ΔT = Q / (m × c)
Substituting the values:
ΔT = 8,280,000 J / (29.5 kg × 4.18 J/g°C)
Simplifying the calculation:
ΔT = 6028.63°C / 29.5
ΔT ≈ 204.37°C
Therefore, the temperature change in Celsius when 29.5L of water absorbs 2.3 kWh of heat is approximately 204.37°C.