- Home
- Standard 11
- Physics
10-1.Thermometry, Thermal Expansion and Calorimetry
medium
A refrigerator converts $500\,g$ of water at $25\,^oC$ into ice at $-10\,^oC$ in $3\,hours\,40\,minutes$ . The quantity of heat removed per minute is ........ $cal/\min$
(Sp. heat of water $1\,cal/gm$, Specific heat of ice $= 0.5\,cal/g\,^oC$ , letent heat of fusion $= 80\,cal/g$ )
A
$100$
B
$150$
C
$200$
D
$250$
Solution
$\frac{\mathrm{Q}}{\mathrm{t}}=\frac{500 \times 25+500 \times 80+500 \times \frac{1}{2} \times 10}{3 \times 60+40}$
$=250 \mathrm{cal} / \mathrm{min}$
Standard 11
Physics