What is the mass of the block of ice?

AI Thread Summary
The discussion centers on calculating the mass of a block of ice that requires 8.8 x 10^5 J of heat to convert from -13°C to water at 13°C. The initial attempt at the solution incorrectly used the heat of fusion value, leading to an erroneous mass calculation of 10.52 kg. After correcting the heat of fusion to 333 x 10^3 J/kg, the correct mass was determined to be 2.11 kg. The importance of careful unit management in calculations was emphasized. Overall, the discussion highlights the significance of accuracy in physics problems.
chrispsu
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im reviewing for my final in a few days an although this is a basic question i can't work out the correct answer for some reason. and i know i started the other thread below this one but i figured this was a different topic so the heading should be different lol

Homework Statement



A heat transfer of 8.8 x 10^5 J is required to convert a block of ice at -13°C to water at 13°C. What was the mass of the block of ice?

Homework Equations



Q=mc(ice)(tf-ti) + mLf + mc(water)(tf-ti)

The Attempt at a Solution



i tried: 8.8x10^5= m(2220)(13) + m(333) + m(4190)(13)
i come out with the 10.52kg which is not correct.
can someone help pinpoint what i am doing something wrong? thanks!
 
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chrispsu said:
i tried: 8.8x10^5= m(2220)(13) + m(333) + m(4190)(13)
Check those numbers. One of them is wrong - it's in the wrong units.

You need to be very careful with your units!
 
oh ok...the heat of fusion should be 333x10^3...solving it out then i would get 2.11 which is the correct answer. I need to be less careless with my units! lol thanks alot! :D
 
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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