A continuous flow of liquid of density 0.8 $g / c m^{3},$ through a pipe has an input electrical power rating = 4 kW and efficiency of conversion of electrical power into heat = 80%. If the liquid is flowing through the device at the rate of 100 cc/sec and the inlet temperature is $10^{0}$ C, the outlet temperature will be[Take specific heat capacity = 1 cal/g] [take 1 cal. Is 4.0 J]

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${13.8}^{0} C$

${22.2}^{0} C$

$20^{0} C$

$16.5^{0} C$

answer is C.

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Detailed Solution

$m s Δ θ = 1600$

$P = \frac{d Q}{d t}$

$\begin{array}{l} \Rightarrow P = m s \frac{d θ}{d t} (∵ d Q = m s d θ) \\ \Rightarrow \frac{80}{100} \times P = \frac{d V}{d t} \times ρ \times s \times d θ \\ \Rightarrow \frac{80}{100} \times 4000 = 100 \times 10^{- 6} \times 0.8 \times 10^{3} \times 4000 \times (θ - 10) \\ \Rightarrow θ - 10 = 10 \\ ∴ θ = 20^{\circ} C \end{array}$