Area under temperature time graph I have an open desiccant heating system which has a flow rate of $0.7$ litres/sec, I have recorded the temperature profile after the air has been heated and plotted it against time in seconds.
My question is how to I work out the total heat added to the air (in kJ) from the graph using $$Q = mC_pdT$$ I know it has something to do with the area under the graph but, I am currently getting confused with how to incorporate the flow rate. Do I work out the total mass of air over the testing time or leave the it as a rate in the equation?
The set-up and temperature profile are shown in the images below. The heat exchanger is to be used in further tests as part of a system which preheats the incoming air. 


 A: Changing the Product Temperature - Heating up with Steam
The amount of heat required to raise the temperature of a substance can be expressed as:
$Q = m C_p dT$         (1)
where
$Q$ = quantity of energy or heat (kJ)
$m$ = mass of the substance (kg)
$C_p$ = specific heat capacity of the substance ($kJ/kg C$ ) - Material Properties and Heat Capacities for several materials
$dT$ = temperature rise of the substance ( C)
Preferring Imperial Units - Use the Units Converter!
This equation can be used to determine a total amount of heat energy for the whole process, but it does not take into account the rate of heat transfer which is:
amount of heat energy per unit time
In non-flow type applications a fixed mass or a single batch of product is heated. In flow type applications the product or fluid is heated when it constantly flows over a heat transfer surface.
Non-flow or Batch Heating
In non-flow type applications the process fluid is kept as a single batch within a tank or vessel. A steam coil or a steam jacket heats the fluid from a low to a high temperature.
The mean rate of heat transfer for such applications can be expressed as:
$q = m C_p dT / t$         (2)
where
$q$ = mean heat transfer rate (kW (kJ/s))
$m$ = mass of the product (kg)
$C_p$ = specific heat capacity of the product (kJ/kg C) - Material Properties and Heat Capacities for several materials
$dT$ = Change in temperature of the fluid ( C)
$t$ = total time over which the heating process occurs (seconds)
Example - Time required to Heat up Water with direct Injection of Steam
The time required to heat 75 kg of water ($C_p = 4.2 kJ/kg C$) from  temperature 20 C to 75 C with steam produced from a boiler with capacity $200 kW (kJ/s)$ can be calculated by transforming eq. 2 to
$t = m C_p dT / q$
$ = (75 kg) (4.2 kJ/kg C) ((75  C) - (20  C)) / (200 kJ/s) $
$ = 86 s$
Note! - when steam is injected directly to the water all the energy from the steam is transferred instantly.
When heating through a heat exchanger - the heat transfer coefficient and temperature difference between the steam and the heated fluid matters. Increasing steam pressure increases temperature - and increases heat transfer. Heat up time is decreased.
Overall steam consumption may increase - due to higher heat loss, or decrease - due to to shorter heat up time, depending on the configuration of the actual system.  
Flow or Continuous Heating Processes
In heat exchangers the product or fluid flow is continuously heated.
The mean heat transfer can be expressed as
$q = C_p dT m / t$         (3)
where
$q$ = mean heat transfer rate (kW (kJ/s))
$m / t$ = mass flow rate of the product (kg/s)
$C_p$ = specific heat capacity of the product (kJ/kg C) - Material Properties and Heat Capacities for several materials
$dT$ = change in temperature of the fluid (C)
Calculating the Amount of Steam
If we know the heat transfer rate - the amount of steam can be calculated:
$ms = q / he$         (4)
where
$ms$ = mass of steam (kg/s)
$q$ = calculated heat transfer (kW)
$he$ = evaporation energy of the steam (kJ/kg)
The evaporation energy at different steam pressures can be found in the SteamTable with SI Units or in the Steam Table with Imperial Units.
Example - Batch Heating by Steam
A quantity of water is heated with steam of 5 bar (6 bar abs) from a temperature of 35 C to 100 C over a period of 20 minutes (1200 seconds). The mass of water is 50 kg and the specific heat capacity of water is 4.19 kJ/kg C.
Heat transfer rate:
$$q = (50 kg) (4.19 kJ/kg C) ((100 C) - (35 C)) / (1200 s)$$
$$    = 11.35 kW$$
Amount of steam:
$$ms = (11.35 kW) / (2085 kJ/kg)$$
$$    = 0.0055 kg/s$$
$$    = 19.6 kg/h$$
Example - Continuously Heating by Steam
Water flowing at a constant rate of 3 l/s is heated from 10 C to 60 C with steam at 8 bar (9 bar abs).
The heat flow rate can be expressed as:
$$q = (4.19 kJ/kg C) ((60 C) - (10 C)) (3 l/s) (1 kg/l)$$
$$    = 628.5 kW$$
The steam flow rate can be expressed as:
$$ms = (628.5 kW) / (2030 kJ/kg)$$
$$    = 0.31 kg/s$$
$$    = 1115 kg/h$$
