Question

What will be the maximum temperature of the maximum if 100 gm ice at $0^\circ C$ is put in 100 gm water at $80^\circ C$. (latent heat = $80\,cal/gm$).
A. $35^\circ C$
B. $45^\circ C$
C. $60^\circ C$
D. $0^\circ C$

Answer 1

We will start answering this question by first writing down all the given values, then we will apply the principle of heat exchange for two objects which will relate mass, specific heat and latent heat of fusion and substitute the given values to find the correct answer to this question.

Formula used:
$m \times {C_{pw}} \times \Delta {T_1} + m \times {\lambda _f} = - \left( {m \times {C_{pw}} \times \Delta {T_2}} \right)$
Where,$M$=mass of water and ice, ${C_{pw}}$= specific heat of water, ${\lambda _f}$= latent heat of fusion and $\Delta {T_1}$ and $\Delta {T_2}$ are changes in temperatures.

Complete step by step answer:
In thermodynamics, a substance's specific heat capacity, also known as massic heat capacity, is the heat capacity of a sample divided by the mass of the sample.One must remember the value of specific heat of water to answer this question.By the principle of heat exchange for two objects at different temperatures kept in isolation,
Heat gained by cold object= Heat lost by hot object

We would assume the system to be isolated and neglecting any heat loss to the surrounding.Given:
$M = 100gm$
$\Rightarrow{C_{pw}} = 4.186Jk{g^{ - 1}}$
$\Rightarrow{\lambda _f} = 334J{g^{ - 1}}$
We can write :
$m \times {C_{pw}} \times \Delta {T_1} + m \times {\lambda _f} = - \left( {m \times {C_{pw}} \times \Delta {T_2}} \right)$

Now lets substitute all the values,
$\Rightarrow 100 \times 4.186 \times \left( {T - 273} \right)100 \times 334 = - \left( {100 \times 4.186 \times \left( {T - 353.14} \right)} \right)$
$\Rightarrow 418.6 \times \left( {T - 273} \right)33400 = - \left( {418.6 \times \left( {T - 353.14} \right)} \right)$
$\Rightarrow 418.6T - 114336.404 + 33400 = - 418.6T + 147824.404$
$\Rightarrow 418.6T + 418.6T = 147824.404 + 114336.404 - 33400$
$\Rightarrow 837.2T = 228760.808$
$\Rightarrow T = 273.2451123\,K$
Now we will convert this into Celsius.Hence:
$\Rightarrow T = 0.1051123^\circ C$
$\therefore T = 0^\circ C$
Hence the final maximum temperature is $T = 0^\circ C$.

Hence the correct answer is option D.

Note: Students make a very common mistake by neglecting the negative sign in the equation: $m \times {C_{pw}} \times \Delta {T_1} + m \times {\lambda _f} = - \left( {m \times {C_{pw}} \times \Delta {T_2}} \right)$. Another common mistake could be that students might stop answering the question right after finding $T = 273.2451123K$. That would be wrong. Remember to convert it into Celsius, because all the options given are in Celsius and not kelvin scale.