Re: Physics help
- From: "V - Man" <Vman@xxxxxxxxxxx>
- Date: Sat, 22 Apr 2006 17:32:51 -0400
"I take 1.0 kg of ice and dump it into 1.0 kg of water and, when
equilibrium
is reached, I have 2.0 kg of ice at 0 deg C. The water was originally at 0
deg C. The specific heat of water = 1.00 kcal/kg*deg C, the specific heat
of
ice = 0.50 kcal/kg*C, and the latent heat of fusion of water is 80
kcal/kg.
The original temp of the ice was.....??????"
The heat given up by the water is
Q_water = L m_water
where L is the heat of fusion and m is the mass.
The heat absorbed by the ice is
Q_ice = C m_ice (T_f - T_i)
where T_i is the initial temperature, and T_f = 0C is the final
temperature.
Is the "C" in the Q_ice formula = to 0.5????? And I assume the m is 1 kg.
I can't imagine the final answer is -160 C for the starting temp of the ice!
Q_water = Q_ice, so, after sorting out the sign convention, solving the
problem becomes simple algebra.
.
- References:
- Physics help
- From: V - Man
- Re: Physics help
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- Re: Physics help
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- Re: Physics help
- From: Gregory L. Hansen
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