Question

A volume of 120ml of drink (half alcohol + half water by mass) originally at a temperature of 25${}^0C$ is cooled by adding 20gm ice at 0${}^0C$. If all the ice melts, find the final temperature (in ${}^0C$) of the drink. (density of drink =0.833gm/cc, specific heat of alcohol =0.6cal/gm ${}^0C$).

Answer 1

To solve this question, we need to know the basic theory and formula related to the specific heat capacity of an object. Here first we will calculate the Mass of drink which is the products of Volume of drink and Density of drink. And then after calculating the final temperature (in ${}^0C$) of the drink using formula as discussed below.

Complete answer:
As given in the question,
Volume of drink = 120ml
Density of drink = 0.833gm/cc
Mass of drink = Volume of drink $\times$ Density of drink
= 120ml$\times$0.833gm/cc
Heat absorbed by ice = 20$\times$80=1600 calories …………… (1)
${d_{drink}} = \dfrac{{{v_{alcohol}} \times {d_{alcohol}} + {v_{water}}{d_{water}}}}{{{v_{alcohol}} + {v_{water}}}}$
0.833 = $\dfrac{{60 \times {d_{alcohol}} + 60 \times 1}}{{60 + 60}}$
${d_{alcohol}}$= $(0.83 \times 2) - 1$
= 0.666
${m_{alcohol}} = 60 \times 0.666 = 40.2$
${m_{water}} = 60 \times 1 = 60$
Heat release by the drink
${Q_{drink}} = \left( {{m_{alcohol}} \times {s_{alcohol}} + {m_{water}} \times {s_{water}}} \right) \times \left( {25 - T} \right)$
= $\left( {40.2 \times O.6 + 60 \times 1} \right)\left( {25 - T} \right)$
= $84.12 \times \left( {25 - T} \right)$ ……………….. (2)
Equating (1) and (2) 1600=84.12×$\left( {25 - T} \right)$
$T = 25 - \dfrac{{1600}}{{84.2}}$
$T = 25 - 19$
T = 6
Therefore, the final temperature (in ${}^0C$) of the drink is 6.

Note:
Heat capacity is the measurable physical quantity that characterizes the amount of heat required to change a substance’s temperature by a given amount. In SI units, heat capacity is expressed in units of joules per kelvin (J/K).