A sample of water with a mass of 648.00 kg at 298 k is heated with 87 kj of energy. the specific heat of water is 1 j-1 kg k-1.
what is the final temperature of the water?
1 answer:
The final temperature of the water is 298. 14 k
<h3>What is quantity of heat?</h3>
The formula for quantity of heat is given as:
Q = mc Δ T = m c ( final temperature - initial temperature)
Where:
- Q is Heat supply = 87 kj
- m is the mass of substance = 648.00 kg
- c is the specific heat capacity = 1 j-1 kg k-1
- ΔT is the temperature change
Substitute the values
87 = 648 × 1 ( T₂ - 298)
87 = 648 ( T₂ - 298)
Divide both sides by 648
87/ 648 = ( T₂ - 298)
0. 134 = T₂ - 298
Make 'T₂' subject of formula
T₂ = 298 + 0. 134
T ₂ = 298. 14 k
Thus, the final temperature of the water is 298. 14 k
Learn more about quantity of heat here:
brainly.com/question/490326
#SPJ1
You might be interested in
It is rounded to the nearest tens. You can only round when you have 1,9. Hope I helped
Answer:
D.
Thats what gauth.math said
What is your grade level and which curriculum are you using?
Answer:
Standard error (S.E) = 0.048 ≅ 0.05
Step-by-step explanation:
<u>Explanation</u>:-
Given sample size 'n'=76
Given the dean who randomly selects 76 students and finds that 58 out of 76 are receiving financial aid.
Sample proportion
The standard error is determined by
Standard error is 0.048 ≅ 0.05
