Heat and temperature change in liquid

AI Thread Summary
To find the mass of portion C needed to achieve a final temperature of 50.0 °C when mixing three portions of the same liquid, the heat lost by portions A and B must equal the heat gained by portion C. The equation Q = cm(Tf - T0) can be applied to each portion, where the specific heat (C) is constant for the same liquid. By setting the heat lost by portions A and B equal to the heat gained by portion C, the relationship can be expressed in terms of mass m and the temperature changes. Rearranging this equation will yield the required mass of portion C in relation to mass m. This approach will provide a clear path to solving the problem.
darw
Messages
3
Reaction score
0

Homework Statement



Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass m and a temperature of 94.0 °C, portion B also has a mass m but a temperature of 78.0 °C, and portion C has a mass mC and a temperature of 34.0 °C. What must be the mass of portion C so that the final temperature Tf of the three-portion mixture is Tf = 50.0 °C? Express your answer in terms of m; for example, mc = 2.20 m.



Homework Equations



C= (m(Tf-T0))/ Q



The Attempt at a Solution



I attempted to fill in the equation for specific heat for each portion and set them equal to each other, because they are the same liquid/same value for C. This just started me going around in circles; I'm clearly going about this the wrong way.

Could anyone provide some guidance to get me started on the right track? Thanks so much for your time.
 
Physics news on Phys.org
sorry, typo.

for the equation I meant: Q = cm (Tf- T0)
 
The heat lost by portion A plus the heat lost by portion B equals the heat gained by portion C if the final temperature is 50 deg C

Write this down as an equation in terms of m, c and the change in temperature for each portion. Rearrange/solve for mC
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top