Question

In a constant flow experiment, a liquid flows at the rate of 0.15kg/min through a tube and is heated dissipating 25.2W. the inflow and outflow temperatures are 15.2 degree Celsius and 17.4 degree Celsius respectively. When the rate of flow is increased to 0.2318kg/min and the rate of heating to 37.8W, the inflow and outflow temperatures remain unaltered. Find the Specific Heat Capacity of the liquid

Answer 1

$mass rate_1 = 0.15kg/min = 0.0025 kg/sec\\ P = 25.2W\\ ∆T = 17.4 - 15.2 = 2.2°C$

$mass\ rate_2 = 0.2318kg/min = 0.00386 kg/sec\\ P = 37.8W \\ ∆T = 17.4 - 15.2 = 2.2°C$

from,

$Pt = mc∆\theta\\ P = \dfrac{mc∆\theta}{t} = \dfrac{m}{t} c∆\theta$

however, Total Power = $\dfrac{m}{t} c∆\theta$ + heat dissipated

$25.2 = 0.0025 × c × 2.2 + H\\ 25.2 = 0.0055c + h ---(i)$

$37.8 = 0.00386 × c × 2.2 + H\\ 37.8 = 0.00849c + h ---(ii)$

subtracting equation (ii) from (i)

$12.6 = 0.00299c\\ c = 4214.05 J/kg.K$