A chunk of ice of mass `m_1 = 100 g` at a temperature `t_1 = 100 g` was at a temperature `t_2`. Assuming the heat capacity of the calorimeter to be negligible, find the entropy increment of the system by the moment the thermal equilibrium is reached. Consider two cases :
(a) `t_2 = 60 ^@C` ,
(b) `t_2 = 94 ^@C`.

Explanation

(a) Here `t_2 = 60^@C`. Suppose the final temperature is `t^@C`. Then
heat lost by water `= m_2 c (t_2 - t)`
heat gained by ice `= m_1 q_m + m_1 c(t - t_1)`, if all ice melts
In this case `m_1 q_m = m_2 xx 4.18(60 - t)`, for `m_1 = m_2`
So the final temperature will be `0^@C` and only some ice will melt.
Then `100 xx 4.18 (60) = m'_1 xx 333`
`m'_1 = 75.3 gm` = amount of ice that will melt
Finally `Delta S = 75.3 xx (333)/(273) + 100 xx 4.18 1n (273)/(333)`
`Delta S = (m'_1 q_m)/(T_1) + m_2 c 1n (T_1)/(T_2)`
=`m_2 c((T_2 - T_1))/(T_1) -m_2 1n (T_2)/(T_1)`
=`m_2 C[(T_2)/(T_1) - 1- 1n(T_2)/(T_1)] = 8.8 J//K`
(b) If `m_2 c t_2 gt m_1 q_m` then all ice will melt as one can check and the final temperature can obtained like this
`m_2 c(T_2 - T) = m_1 q_m + m_1 c(T - T_1)`
`(m_2 T_2 + m_1 T_1) c - m_1 q_m = (m_1 + m_2) cT`
or `T = (m_2 T_2 + m_1 T_1 - (m_1 q_m)/(c))/(m_1 +m_2) ~= 280 K`
and `Delta S = (m_1 q)/(T_1) + c(m_1 1n (T)/(T_1) -m_2 1n (T_2)/(T)) = 19 J//K`.