How Do You Calculate the Mass of an Aluminum Cup in a Heat Transfer Problem?

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To calculate the mass of the aluminum cup in this heat transfer problem, use the principle that heat lost equals heat gained. The equation Q = mCΔT applies to each component: the water, silver sample, copper stirrer, and aluminum cup. Each object will reach the final equilibrium temperature of 25.7°C, with the water and stirrer starting at 21.2°C. The silver, initially at 80.8°C, will lose heat, while the other components will gain heat. Setting up the equation correctly will allow for the determination of the aluminum cup's mass.
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Homework Statement


An aluminum cup contains 217g of water at 21.2oC. A 404g sample of silver at an initial temperature of 80.8oC is placed in the water. A 41.0g copper stirrer is used to stir the mixture until it reaches its final equilibrium temperature of 25.7oC. Calculate the mass of the aluminum cup.



Homework Equations


Q=mCdeltaT
heat lost=heat gained


The Attempt at a Solution


Could someone please help me set the problem up w/o numbers. Thanks.
 
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You have the right equations to work with, maybe listing a few facts will help get you going:

Every object will end up at the final temperature of 25.7C.

Each object either loses heat, or gains heat. That tells you on which side of the heat lost=heat gained equation each object belongs.

Presumably the copper stirrer also starts out at 21.2C.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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