Question

The amount of heat (in calories) required to convert 10g of ice at -${10}^{0}C$ into steam at ${100}^{0}C$ is nearly:

(Specific heat of steam is $536cal/gm-K$)

• A. $5400cal$
• B. $6400cal$
• C. $7210cal$
• D. $8250cal$

Answer 1

The correct option is C $7210cal$

Heat required $=$$-$10${}^0$C ice to 0${}^0$C ice + 0${}^0$C ice to 0${}^0$C water + 0${}^0$C water to 100${}^0$C water + 100${}^0$C water to 100${}^0$C steam
$=10\left[0.487\right(10)+80+1(100-0)+536]$

$=7208.7cal$ [since specific heat of ice$=$0.487cal/g-K]