Can someone please help me solve this homework problem for my chemistry class.Thanks
A 4.0*10^1kg sample of water absorbs 308KJ of heat. If the water was initially at 25.5 C, what is its final temperature?
To solve this problem, we can use the equation:
q = mcΔT
where q is the heat absorbed by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
We are given:
- q = 308 kJ
- m = 4.0 * 10^1 kg
- initial temperature (T1) = 25.5 °C
The specific heat capacity of water (c) is approximately 4.18 J/g°C, but since we are given the mass in kg, we need to convert it into grams.
First, we convert the mass into grams:
4.0 * 10^1 kg = 4.0 * 10^4 g
Next, we rearrange the equation to solve for ΔT:
q = mcΔT
ΔT = q / mc
Now, plug in the given values:
ΔT = (308,000 J) / (4.0 * 10^4 g * 4.18 J/g°C)
Perform the calculation:
ΔT = 1.47 °C
Finally, to find the final temperature (Tf), we add the change in temperature (ΔT) to the initial temperature (T1):
Tf = T1 + ΔT
Tf = 25.5 °C + 1.47 °C
Calculate the final temperature:
Tf ≈ 26.97 °C
Therefore, the final temperature of the water is approximately 26.97 °C.
Sure, I can help you solve this problem. To find the final temperature of the water, we can use the equation:
q = m * c * ΔT
where:
q represents heat transfer (in this case, 308 KJ),
m represents mass (4.0 * 10^1 kg),
c represents specific heat capacity of water (4.18 J/g°C),
ΔT represents the change in temperature.
To solve for ΔT, we need to rearrange the equation:
ΔT = q / (m * c)
Now, let's plug in the values we have:
4.0 * 10^1 kg = 40 kg
q = 308,000 J (converted from 308 KJ)
c = 4.18 J/g°C
Now we can calculate:
ΔT = 308,000 J / (40 kg * 4.18 J/g°C)
To proceed with the calculation, we need to convert grams to kilograms:
1 g = 0.001 kg
So, the equation becomes:
ΔT = 308,000 J / (40,000 g * 4.18 J/g°C)
Simplifying further:
ΔT = 308,000 J / 167,200 J/°C
Now, divide:
ΔT = 1.842°C
Finally, we can find the final temperature by adding the change in temperature to the initial temperature:
T(final) = 25.5°C + 1.842°C
Therefore, the final temperature of the water is approximately 27.342°C.