A) A gas-fired heat exchanger uses gas with energy value of 40MJ / m^3 at a rate of 0.4 m^3/h and produces 50kg/h of water at a temperature of 70^oC from feed water at 20^oC. The rest of the question is in the picture attached.PLZ help?

Let us calculate amount of heat required to raise the temperature of feed water at #20^@C# to water at #70^@C# by the gas fired heat exchanger in one hour.

Mass of water used in one hour #m=50\ kg#
Rise in temperature #DeltaT=70^@-20^@=50^@C#
Useful heat #Q_u=msDeltaT#

#=>Q_u=50xx(4.19xx10^3)xx50#
#=>Q_u=10475000\ J#

Heat produced by the gas in the heat exchanger in one hour

#Q_p=0.4xx(40xx10^6)#
#=>Q_p=16000000\ J#

Thermal efficiency of the heat exchanger #eta=Q_u/Q_p#

#eta=10475000/16000000=0.65#, rounded to two decimal places.

Start by working out how much energy is required to heat 50 kg of water from #20^oC# to #70^oC# (1 hr worth):

Specific heat capacity of water = #4.19(kJ)/(kg^oC)#

so 50 kg raised 50 degrees will need

Energy #= 4.19(kJ)/cancel(kg^oC) * 50cancel(kg) *50^ocancelC #

# = 10475 kJ#

# = 10.475 MJ# This is the energy/hr transferred to the water

Energy/hr in from gas # = 40(MJ)/cancelm^3 * 0.4 cancelm^3 = 16MJ#

so thermal efficiency is #("heat out")/("heat in")#

# = (10.475 MJ)/(16MJ)#

# = 0.6546875#

# = 65.47#% (2dp)