What will be the final temperature of a 655g sample of water, initially at 17.0oC, after 77.8 kJ have been added to it?
@CShrix
Use formula of specific heat capacity\[E=m*c*(T _{f}-T _{i})\] \[77800=0.655*4.18*(T _{f}-17)\] Find Tf.
lol "tf"
77800=-46.5443Tf
@LeibyStrauss Tf=-1671.525837
It doesn't look right. The initial T was 17 degree celsius. If heat was added you won't have a negative number. How did you get -46.5443Tf If you use the formula written by Hoslos you need to distribute 655∗4.18 to Tf as well as to 17 As a correction q = m*c*(Tfinal - T initial) mass is 655 not 0.655 Do you have the formula q = m*c*(Tfinal - T initial) or do you have a similar formula? If you have the same one how can you re-write (Tfinal - Tinitial)?
I do not.
What formula do you have for q? q = heat
No formulas were given on the sheet.
q = mass * specific heat capacity * change in temperature There are different ways of writing the formula, one way is \[q = m \times c \Delta T\] q = heat m = mass in grams c = specific heat capacity J/gK Delta T = T final - T initial in Kelvin Can you plug in the values in to the formula?
So I need to change the 17.0 oC to K?
No. Since it is change of T. Meaning, if something changes from 10 degrees Celsius to 25 degrees celsius, the change in T is 15 degrees Celsius. If we convert it to K it is 288.15K - 283.15K which is 15 K. So we have the same change whether it is C or K.
I'm getting confused
For now just plug in the values you have into the formula. We can come back later to your question regarding converting to K. For now, you can solve the problem as if the units of the formula are in C
655g * 77.8* 17.0oC
Is that it basically?
q = m*c*deltaT The units on q is J Your question has 77.8 kJ
@staldk3 , I made a slim mistake on my workout. Instead of converting the mass to kg, it should remain as g. Hence being:\[\frac{ 77800 }{ 655*4.18 }=T _{f}-17\]\[\frac{ 77800 }{ 655*4.18 }+17=45.4 C\]
Tf= 45.4 Degrees.
So the kJ has to be converted to J which makes it 77800 J?
Correct
Finally got the right answer.
Are you done?
And we need 3 sig figs instead of 2?
Correct.
Why is that? What about the +17?
The original question had how many sig figs?
Oh okay so you just left out the .0 part?
17.0 is 3 sig figs
Right. But in the equation it was left as 17. Why I thought there should be 2 sig figs.
I did not post the equation. If you want to work it through step by step, I can help you, but you need to plug the values in.
I did it on a calculator and got the same answer. I'm good on it. Just got mixed up on the 17 part for a bit but I think they just left off the .0 on the 17.0 just to type it easier. It's all good.
Ok.
Thanks
You're welcome
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