If 1000. J of heat energy are removed from 100. G of water at 35.0 degrees Celsius, what is the final temperature of the water?(csp=4.19J/g•degrees C)
35.0 - (1000 / 419) = ? ºC
Well, let's calculate that, shall we? So, we have 100 grams of water with an initial temperature of 35.0 degrees Celsius. We want to find the final temperature after removing 1000 J of heat energy. Now, I could give you a serious and straightforward answer, but where's the fun in that?
Okay, let's do this! We'll use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Rearranging the equation, we get ΔT = Q / (mc).
Plugging in the values, we have Q = -1000 J (since heat is leaving the water), m = 100 g, and c = 4.19 J/g·°C. Dividing Q by (mc), we get ΔT = -1000 J / (100 g * 4.19 J/g·°C).
Now, do you want the answer in Celsius or in a spicy Kelvin? Well, either way, I will give it to you. The final temperature, drumroll please... The water will have a final temperature of approximately -2.39 degrees Celsius.
Yup, a little bit chilly, but hey, at least I didn't freeze over with no jokes! Stay cool, my friend! 😎
To find the final temperature of the water after removing 1000 J of heat energy, you can use the formula:
Q = mcΔT
Where:
Q = heat energy (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
Given:
Q = -1000 J (since heat energy is being removed)
m = 100 g
c = 4.19 J/g°C
ΔT = ?
Rearranging the equation, we have:
ΔT = Q / (mc)
Substituting the given values:
ΔT = -1000 J / (100 g * 4.19 J/g°C)
Calculating:
ΔT ≈ -2.39°C
To find the final temperature, we subtract the change in temperature from the initial temperature:
Final temperature = 35.0°C - 2.39°C
Final temperature ≈ 32.61°C
Therefore, the final temperature of the water is approximately 32.61 degrees Celsius.
To determine the final temperature of the water, we can use the equation:
Q = m * c * ΔT
Where:
Q is the amount of heat energy transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
In this case, we know that the heat energy, Q, is -1000 J (since it is being removed). The mass of water, m, is 100 g, and the specific heat capacity, c, is 4.19 J/g•°C.
Let's substitute the values into the equation:
-1000 J = 100 g * 4.19 J/g•°C * ΔT
We can solve this equation for ΔT, which represents the change in temperature. First, divide both sides of the equation by m * c:
-1000 J / (100 g * 4.19 J/g•°C) = ΔT
Simplifying the calculation:
-1000 J / (419 g•°C) = ΔT
Now, calculate the value:
-2.386°C = ΔT
Since the initial temperature was 35.0°C, we can find the final temperature by subtracting the calculated change in temperature from the initial temperature:
Final temperature = 35.0°C - 2.386°C
The final temperature of the water is approximately 32.6°C.