Gujarati
Hindi
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

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