Question

A $10$ gm ice cube is dropped into $45$ gm of water kept in a glass. If the water was initially at a temperature of ${28}^{o}C$ and the temperature of ice $-{15}^{o}C$, find the final temperature (in ${}^{o}C$) of water. (Specific heat of $ice=0.5cal/gm{/}^{o}C$ and $L=80cal/gm$)

• A. $T_f=5^{o}C$
• B. $T_f=7^{o}C$
• C. $T_f=9^{o}C$
• D. $T_f={12}^{o}C$

Answer 1

The correct option is B $T_f=7^{o}C$

Given - mass of ice , $m_i=10g$ ,

mass of water , $m_w=45g$ ,

initial temperature of ice , $t_i=-{15}^{o}C$ ,

initial temperature of water , $t_w={28}^{o}C$ ,

specific heat of ice , $c_i=0.5cal/g{-}^{o}C$ ,

latent heat of ice , $L=80cal/g$ ,

when the ice is dropped into water , water being at a higher temperature , supplies heat to ice . Ice , by getting heat from water , first reaches to $0^{o}C$ , then melts at $0^{o}C$ , and then temperature of this water at $0^{o}C$ increases to a final temperature (let $t_f$) .

Hence , total heat taken by ice ,

$Q=m_i{c}_i(0+15)+m_{i}L+m_i{c}_w(t_f-0)$ ,

total heat given by water ,

$Q^{\prime }=m_w{c}_w(28-t_f)$ ,

from the principle of mixtures ,

heat taken =heat given ,

$m_i{c}_i(0+15)+m_{i}L+m_i{c}_w(t_f-0)$=$m_w{c}_w(28-t_f)$ ,

or $10\times 0.5\times 15+10\times 80+10\times 1\times t_f=45\times 1\times (28-t_f)$ ,

or $75+800+10t_f=1260-45t_f$ ,

or $55t_f=1260-875=385$ ,

or $t_f=385/55=7^{o}C$