Adding heat for final temperature

AI Thread Summary
To calculate the heat required to raise the temperature of 0.250 kg of water from 20.0°C to 30.0°C, the correct specific heat value of water should be used, which is 4186 J/kg·K. The initial calculation mistakenly used a specific heat of 1, leading to an incorrect result of 10.4725. The correct heat required is 1.05 x 10^4 J. For the mercury, using its specific heat of 0.140 J/g·K, the final temperature can be calculated by rearranging the heat equation. Accurate unit conversion and specific heat values are crucial for obtaining the correct results.
shakejuhn
Messages
17
Reaction score
0

Homework Statement


How much heat is required to raise the temperature of 0.250kg of water from 20.0oC to 30.0oC?

If this amount of heat is added to an equal mass of mercury that is initially at 20.0oC, what is its final temperature?


Homework Equations



Q=c*M*(t2-t1)

The Attempt at a Solution



for the first part i got
q=1*0.250*(30.0-20.0)
q=10.4725
this is wrong the correct answer was 1.05*10^4
how did they get that answer?

The second part i have so far
1.05*10^4=c*0.250*(t2-20)
it didnt say what the spefice heat of mercury was so i looked it up and it was 0.140
1.05*10^4=0.140*0.250*(t2-20)
1.05*10^4/(.035)+20

please help with this
 
Physics news on Phys.org
For the first part why are you using 1 for c? Look up the specific heat of water. Regardless, I don't see how you got 10.4725 from your equation for q. Is there a typo in there somewhere?
 
shakejuhn said:

Homework Statement


How much heat is required to raise the temperature of 0.250kg of water from 20.0oC to 30.0oC?

If this amount of heat is added to an equal mass of mercury that is initially at 20.0oC, what is its final temperature?


Homework Equations



Q=c*M*(t2-t1)

The Attempt at a Solution



for the first part i got
q=1*0.250*(30.0-20.0)
q=10.4725
this is wrong the correct answer was 1.05*10^4
how did they get that answer?

The second part i have so far
1.05*10^4=c*0.250*(t2-20)
it didnt say what the spefice heat of mercury was so i looked it up and it was 0.140
1.05*10^4=0.140*0.250*(t2-20)
1.05*10^4/(.035)+20

please help with this

You have to be clear on your units. You appear to be working in calories and kg. If so, you have to use a specific heat for water in calories/kg deg not calories/g deg.

AM
 
i just said find the answer in joules
 
shakejuhn said:
i just said find the answer in joules
If you are using a specific heat of 1 for water you are working in calories/g. Kelvin. If you want the answer in joules you have to multiply the answer by 4.186 (ie. 4.186 J in one cal.). The specific heat for water in Joules/Kg K is 4186.

AM
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top