Re: Physics help

In article <e2ditp$8cp7$1@xxxxxxxxxxxxxxxxx>, V - Man <Vman@xxxxxxxxxxx> wrote:

Most folks here are OK with providing help as long as you show your
attempt at the problem.

Thanks for replying, Phil. Unfortunately, my attempt is futile. I have no
idea how to approach solving this problem. Our prof. is a chemistry prof.
and has never taught physics before. So, I am trying with no success to
answer this one question based on his limited lecture notes. The text is not
much of a help either. All we get are the formulas with no problems to
practice putting the formulas to work!!

Here's the problem (if anyone can help, great! If not, I understand):

"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.....??????"

If anyone would just maybe be kind enough to even help setting it up, I
would really appreciate it. Thanks and again, sorry for the inrusion.

As John Christiansen noted, the temperature of the water does not change.
Another thing to note is that none of the original chunk of ice melts.

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.

Q_water = Q_ice, so, after sorting out the sign convention, solving the
problem becomes simple algebra.

--
"He who only sees business in business is a fool."
.

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