If 1.68 L of water is initially at 25.0 °C, what will its temperature be after absorption of 6.8 10-2 kWh of heat?
I don't know how to set it up because i have the equation q= m times c times change in temp (the triangle and the a T is the symbol for that. I know ther is a specific name but i forget it at the moment.)
thanks:)
The triangle is a capital Greek letter delta.
The equation you have is the correct one to use. What is the problem? You simply use direct substitution of the numbers into the equation.
I got it:)
Great!
No problem, I can help you with that!
To solve this problem, you can use the equation q = mcΔT, where q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
First, you need to calculate the amount of heat absorbed by the water. You have the energy in kilowatt-hours (kWh), but you need to convert it to joules since the specific heat capacity of water is usually given in joules per gram per degree Celsius (J/g°C).
To convert kilowatt-hours to joules, use the conversion factor 1 kWh = 3.6 x 10^6 J.
Now, you can calculate the amount of heat absorbed:
q = (6.8 x 10^-2 kWh) x (3.6 x 10^6 J/1 kWh)
= 2.448 x 10^5 J
Next, we need to calculate the mass of water. We know the volume of water is 1.68 L, but we need to convert it to grams since the specific heat capacity of water is usually given in J/g°C. The density of water at room temperature is approximately 1 g/mL. Therefore:
mass = volume x density
= 1.68 L x 1 g/mL
= 1.68 kg (since 1 mL = 1 g)
Now, we can calculate the change in temperature (ΔT) using the equation q = mcΔT. Rearranging the equation gives:
ΔT = q / (mc)
Using the values we have determined:
ΔT = (2.448 x 10^5 J) / [(1.68 kg) x (4.18 J/g°C)]
≈ 35.44 °C
Finally, to find the final temperature, add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
= 25.0 °C + 35.44 °C
= 60.44 °C
Therefore, the final temperature of the water will be approximately 60.44 °C after absorbing 6.8 x 10^-2 kWh of heat.