To increase the temperature of 100.0 g of ice from -50.0? to -10.0?, how
much energy is required?

We are given:

Mass of ice = 100 grams

Initial temperature = -50°C

Final Temperature = -10°C

We know that the specific heat of ice is 2.09 joules / (gram * °c)

Heat Absorbed:

We know that:

ΔQ = mcΔT                                                    [where ΔQ is the heat absorbed]

ΔQ = (100 grams)(2.09)(-10 - (-50))

ΔQ = (100)(2.09)(40)

ΔQ = 8360 Joules

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.