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Standard X

Physics

Factors Affecting Resistance & How They Affect

An electrical...

Question

An electrically heating coil is placed in a calorimeter containing 360 g of H $_{2}$ O at 10 $^{o}$ C. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and the coil is 40 g. The temperature of water after 10 minutes will be

{42.15}^{0} C

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{32.14}^{0} C

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{22.14}^{0} C

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{52.14}^{0} C

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Solution

The correct option is A ${42.15}^{0} C$
Heat produced by the coil
$Q = P . t = (m + W) c Δ T$
$\frac{90 \times 600}{4.2} = (360 + 40) Δ T$
Thus we get the temperature difference as
$Δ T = {32.14}^{0} C$
or $T = 10 + 32.14 = {42.14}^{0} C$

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An electrically heating coil is placed in a calorimeter containing 360 g of H2O at 10oC. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and the coil is 40 g. The temperature of water after 10 minutes will be

The correct option is A 42.150CHeat produced by the coilQ=P.t=(m+W)cΔT90×6004.2=(360+40)ΔTThus we get the temperature difference asΔT=32.140Cor T=10+32.14=42.140C

An electrically heating coil is placed in a calorimeter containing 360 g of H $_{2}$ O at 10 $^{o}$ C. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and the coil is 40 g. The temperature of water after 10 minutes will be

The correct option is A ${42.15}^{0} C$
Heat produced by the coil
$Q = P . t = (m + W) c Δ T$
$\frac{90 \times 600}{4.2} = (360 + 40) Δ T$
Thus we get the temperature difference as
$Δ T = {32.14}^{0} C$
or $T = 10 + 32.14 = {42.14}^{0} C$