Temperature & Heat: Equilibrium of 1 kg Aluminium & Water

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To find the equilibrium temperature when 0.1 kg of aluminum at 90°C is submerged in 1 kg of water at 10°C, the heat lost by aluminum must equal the heat gained by water. The specific heat capacities are 920 J/(kg·°C) for aluminum and 4186 J/(kg·°C) for water. Setting the heat transfer equations for both substances equal allows for solving the two unknown temperature changes, ΔT. The definition of ΔT is crucial, as it is the difference between the final and initial temperatures. Ultimately, the equilibrium temperature can be calculated using these principles.
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A .1 kg piece of aluminum at 90 degrees C is submerged in 1 KG of water at 10 degrees C. What is the temperature when they reach equalibrium?

Now I know, aluminum has a c of 920 J/(kg * degree C) and water has a c of 4186 J/(kg * degree C).

I'm not sure how to approach this.

Best I can think of is setting m*c*dT for the water equal to m*c*dT for the aluminum. But I have two unknown variables there. (The dT's) So that can't be right. Help is appreciated.
 
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What is the definition for your delta T? It is \Delta T = T_{end} - T_{start}

Does that point you in the right direction? Also, think about the physical situation when the two have reached equillibrium...
 
superdave said:
A .1 kg piece of aluminum at 90 degrees C is submerged in 1 KG of water at 10 degrees C. What is the temperature when they reach equalibrium?

Now I know, aluminum has a c of 920 J/(kg * degree C) and water has a c of 4186 J/(kg * degree C).

I'm not sure how to approach this.

Best I can think of is setting m*c*dT for the water equal to m*c*dT for the aluminum. But I have two unknown variables there. (The dT's) So that can't be right. Help is appreciated.
This means you will need to work out two equations to determine the values of these two different delta T's.

AM
 
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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