Question

A fuel oil heater receives 23200 lb of oil per hour at 80°F and discharges it at 160°F. The specific heat of the oil is 0.48 BTU/lb-°F. Find the rate of heat transfer to the oil, BTU/hr.

Answer 1

As per the given question,

quantity of flue=23200 lb

$T_1=80^oF$

$T_2=160°F.$

specific heat of the oil is $(s)=$ 0.48 BTU/lb-°F

we know that heat, $Q=ms(T_2-T_1)$

$Q=232200\times 0.48\times (160-80)=8916480BTU$

Hence rate of heat flow per hour $\dfrac{dQ}{dt}=8916480BTU/hour$