Question

A quantity of ice ( 5 kg) at (0 C) was added to a quantity of water ( 1 kg) at (35 C). By neglecting the amount of heat lost. Calculate the final temperate of the mixture.

Given that:
Cw = 4180 J/kg.k
Lf = 3.33* 10^5 j/kg

Answer 1

qualitatively, we can say that the melting of the sample of ice occurs, subsequently, the corresponding water will be heated to the final temperature, namelt the equilibrium temperature \(t_e\). So we can write the subsequent equation of energy balance:
\[\huge \begin{gathered}
{L_f}{M_1} + {c_{{H_2}O}}{M_1}\left( {{t_e} - {t_0}} \right) = \hfill \\
\hfill \\
= {c_{{H_2}O}}{M_{water}}\left( {{t_2} - {t_e}} \right) \hfill \\
\end{gathered} \]
where \(\Large t_2=35°C\), \(\Large t_0=0°C\), and \(\Large c_{H_2O}=1J/(Kg\cdot K)\)
Furthermore, \(\Large M_1\) is the mass of the initial sample of ice

Answer 2

Please solve that formula for \(\Large t_e\)

Answer 3

@Michele_Laino , I want you to solve it totally by yourself. As I am going to discuss with you something related to the final answer.

Answer 4

I got a negative result as final temperature, maybe there is a typo among your data

Answer 5

No, that's correct, You got -60 ?

Answer 6

@Michele_Laino

Answer 7

yes! you are right! There is no reason to use the specific heat of ice

Answer 8

Now explain what does -60 means.

Answer 9

I studied negative temperatures, when I have to study statistical physics, so please wait, I go to retrieve my textbook of statistical physics

Answer 10

Okie dokie !

Answer 11

Thanks for your valuable information, time.

Answer 12

please wait, I'm thinking on your question...

Answer 13

@Michele_Laino , I closed as I though you would retrieve your textbook first.

Whenever you find the answer just tag.