Specific Heat Problem and Thermos

AI Thread Summary
An insulated Thermos contains 127 cm³ of coffee at 79°C, and a 12.0 g ice cube is added to cool it. The problem involves calculating the temperature change of the coffee after the ice melts and thermal equilibrium is reached, treating both substances as pure water. The total heat exchange equation is set to zero, combining the heat absorbed by the melting ice and the heat lost by the coffee. The calculations yield a final temperature of approximately 65.3°C for the coffee, resulting in a cooling of about 13.7°C. The discussion emphasizes breaking the problem into two steps: the melting of the ice and the subsequent heating of the melted ice.
RTucekr
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Homework Statement


An insulated Thermos contains 127 cm3 of hot coffee at 79°. You put a 12.0 g ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

mw=0.127kg
Tiw=79+273=352K
mi=0.012kg
Tii=273K

Homework Equations


Q=mcΔT
Q=mLf

The Attempt at a Solution


Qtotal=Qwater+Qice melting+Qice warming=0

mILf+mIcw(Tf-Ti)+mwcw(Tf-Ti)=0
(0.012kg)(333x10^3 J/kg) + (0.012kg)(4180 J/(kg*k))(Tf-273K)+(0.127kg)(4180 J/(kg*K))(Tf-352)=0
Tf=338.303K or 65.302 C

ΔT=79-65.302=13.697C

Any help would be appreciated. I have no idea where I am going wrong with this one.
 
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Break it up into two steps. M= mass of water and m = mass of ice.

1. Mass m of ice melts, absorbing mL units of energy, where L is the latent heat of fusion.
Temp of mass M of water drops to certain t1 from 79 C by giving out mL amount of heat. t1 can be found.

2. Mass m of water heats up from 0 C to t2, and mass M of water cools down to t2 from t1. Find t2.

If you use cgs units for this one, take 80 cal/g for L.
 
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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