Can someone please help me solve this homework problem for my chemistry class.It's my last problem and I'm stuck!:/ 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?
q = mass x specific heat x (Tfinal-Tinitial)
q = 308,000 J
mass = 40000
To solve this problem, we can use the equation
Q = m * c * ΔT
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 that the heat absorbed by the water (Q) is 308 KJ, the mass of the water (m) is 4.0 * 10^1 kg, and the initial temperature (T1) is 25.5 C. We need to find the final temperature (T2).
The specific heat capacity of water (c) is approximately 4.18 J/g°C.
First, let's convert the mass of the water from kg to grams:
4.0 * 10^1 kg = 4.0 * 10^4 g
Now, we can rewrite the equation as:
Q = m * c * ΔT
308 KJ = (4.0 * 10^4 g) * (4.18 J/g°C) * (T2 - 25.5 C)
To solve for T2, we can rearrange the equation:
T2 = (Q / (m * c)) + 25.5 C
Now, substitute the given values into the equation:
T2 = (308 * 10^3 J) / ((4.0 * 10^4 g) * (4.18 J/g°C)) + 25.5 C
After calculating this expression, you will get the final temperature (T2) in degrees Celsius.