Help Needed: Solving Tf in Metal & Liquid Heat Exchange

AI Thread Summary
The discussion revolves around solving a heat exchange problem involving a metal and a liquid. The user initially sets up the energy conservation equation incorrectly by omitting a negative sign, which is crucial for indicating heat loss from the metal and heat gain by the liquid. After receiving feedback, the user acknowledges the mistake and thanks the respondent for the clarification. The conversation highlights the importance of correctly applying the principles of thermodynamics in heat transfer problems. Understanding energy conservation is key to solving such equations accurately.
tedjj
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I can't solve this problem. Could anyone help me?
A piece of metal with a mass of 2kg and specific heat of 200J/kg C is initially at a temperature of 120 C. The metal is placed into an insulated container that contains a liquid of mass 4kg, a specific heat of 600J/kg C, and an initial temperature of 20C.
I am using equation
Mm * Cm(Tf-Ti)=Ml*Cl(Tf-Ti)
2*200*(Tf-120)=4*600*(Tf-20)
(Tf-120)=6*(Tf-20)
Tf-120 = 6Tf-120
Tf=6Tf
Where did I make a mistake?
Thanks
 
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In the first formula you use there should be a minus sign on one of the sides. This is an energy equation; energy is conserved so the amount of heat that the metal loses (-) is gained (+) by the liquid...
 
ohh You are right. Thank you very much.
 
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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