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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

At 1 atmospheric pressure, 1 gm of water having a volume of $1\: cm^3$ becomes $1.091\: cm^3$ of ice on freezing. The heat of fusion of water at 1 atmosphere is 80 cal/g. What is the change in internal energy during the process?

$\begin {array} {1 1} (A)\;80.0022 \: cal & \quad (B)\;79.9978 \: cal \\ (C)\;-80.0022 \: cal & \quad (D)\;-79.9978 \: cal \end {array}$

1 Answer

Heat given out during freezing
$Q = -mL = -1 \times 80 = -80\: cal$
External work done, $W = P(V_{ice} – V_{water} ) = 1.013 \times10^5 \times (1.091 – 1) \times 10^{-6}$
$= 9.22 \times 10^{-3} J = 9.22 \times 10^{-3} / 4.18\: cal = 0.0022 \: cal$
Thus, change in internal energy
$ U = Q – W = - 80 – 0.0022 = -80.0022\: cal$
Ans : (C)
answered Mar 15, 2014 by thanvigandhi_1
 

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