# Heat and temperature change in liquid

1. Dec 6, 2011

### darw

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

2. Dec 6, 2011

### darw

sorry, typo.

for the equation I meant: Q = cm (Tf- T0)

3. Dec 6, 2011

### Stonebridge

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