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10-1.Thermometry, Thermal Expansion and Calorimetry
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Find the quantity of heat required to convert $40\; gm$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L _{\text {ice }}$ $=0.336 \times 10^6 J / kg$.
specific heat of ice $=2100 \;J / kg - K$ sp heat of water= $4200\; J / kg - K$
A
$18480$
B
$17580$
C
$18685$
D
$17412$
Solution
(d) Mass of ice $=\frac{40}{1000}=0.04 Kg$
Heat required to raise the temperature of ice from $-20^{\circ} C$ to $0^{\circ} C =0.04 \times 2100 \times 20=1680 J$
Heat required to convert the ice into water at $0^{\circ} C = mL$
$=.04 \times 0.336 \times 10^6=13440 J$
Heat required to heat water from $0^{\circ} C$ to $20^{\circ} C$
$=.04 \times 4200 \times 20=3360 J$
Total heat required $=1680+13440+3360=18480\; J$
Standard 11
Physics
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