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Standard XII

Chemistry

First Law of Thermodynamics

A refrigerato...

Question

A refrigerator converts $100 g$ of water at $25^{o} C$ into ice at $- 10^{o} C$ one hour and $50$ minutes. The quantity of heat removed per minute is:-( Specific heat of $i c e = 0.5 c a l / g^{o} C$ , latent heat of fusion $= 80 c a l / g)$ .

50 c a l

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100 c a l

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200 c a l

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75 c a l

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Solution

The correct option is B $100 c a l$
$H = m C P_{w} Δ T_{1} + m L + m C P_{i} Δ T_{2}$
$= m [C P_{w} Δ T_{1} + L + C P_{i} Δ T_{2}]$
$= 100 [1 \times (25 - 0) + 80 + 0.5 \times (0 - (- 10))]$
$= 100 [25 + 80 + (0.5 \times 10)]$
$= 100 [25 + 60 + 5]$
$= 11$ K Cal
Heat removed in $1$ hr $50$ min = $11 \times 10^{3}$ Cal
$1$ hr $50$ min $= (60 + 50)$
$= 110$ min
$110$ min $= 11 \times 10^{3}$ Cal
$1$ min $= \frac{11 \times 10^{3}}{110}$
$= 100$ Cal

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Similar questions

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A refrigerator converts 100 g of water at 25oC into ice at −10oC one hour and 50minutes. The quantity of heat removed per minute is:-( Specific heat of ice=0.5 cal/goC, latent heat of fusion =80 cal/g).

The correct option is B 100 calH=mCPwΔT1+mL+mCPiΔT2 =m[CPwΔT1+L+CPiΔT2] =100[1×(25−0)+80+0.5×(0−(−10))] =100[25+80+(0.5×10)] =100[25+60+5] =11 K CalHeat removed in 1 hr 50 min = 11×103 Cal1 hr 50 min =(60+50) =110 min 110 min =11×103 Cal1 min =11×103110 =100 Cal

A refrigerator converts $100 g$ of water at $25^{o} C$ into ice at $- 10^{o} C$ one hour and $50$ minutes. The quantity of heat removed per minute is:-( Specific heat of $i c e = 0.5 c a l / g^{o} C$ , latent heat of fusion $= 80 c a l / g)$ .