Question
$$2\,kg$$ of ice at $$ - {20^ \circ }C$$ is mixed with $$5\,kg$$ of water at $${20^ \circ }C$$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are $$1\,k\,cal/kg/{\,^ \circ }C\,\,\& \,\,0.5\,k\,cal/kg/{\,^ \circ }C$$ while the latent heat of fusion of ice is $$80\,k\,cal/kg$$
A.
$$7\,kg$$
B.
$$6\,kg$$
C.
$$4\,kg$$
D.
$$2\,kg$$
Answer :
$$6\,kg$$
Solution :
Heat required to convert $$5\,kg$$ of water at $${20^ \circ }C$$ to $$5\,kg$$ of water at $${0^ \circ }C$$
$$ = m{C_\omega }\Delta T = 5 \times 1 \times 20 = 100\,k\,cal.$$
Heat released by $$2\,kg.$$ Ice at $$ - {20^ \circ }C$$ to convert into $$2\,kg$$ of ice at $${0^ \circ }C$$
$$ = m{C_{{\text{ice}}}}\Delta T = 2 \times 0.5 \times 20 = 20\,k\,cal.$$
How much ice at $${0^ \circ }C$$ will convert into water at $${0^ \circ }C$$ for giving another $$80\,k\,cal$$ of heat
$$\eqalign{
& Q = mL \cr
& \Rightarrow \,\,80 = m \times 80 \cr
& \Rightarrow \,\,m = 1\,kg \cr} $$
Therefore the amount of water at $${0^ \circ }C$$
$$= 5\,kg + 1\,kg = 6\,kg$$
Thus, at equilibrium, we have, [$$6\,kg$$ water at $${0^ \circ }C + 1\,kg$$ ice at $${0^ \circ }C$$ ].