# Answer to Question #5464 in Molecular Physics | Thermodynamics for Mackenzie

Question #5464

How many kJ are needed to convert 14.4 grams of ice at -80.0 degrees Celsius to water at 61 degrees Celsius?

Expert's answer

E1 = Ci*(0-(-80))*0.0144 = 1800*80*0.0144 = 2430.72 J,

where Ci = 1800 J/kg*K is the average specific heat of an ice (from -80 to 0 degree).

E2 = Cmi*0.0144 = 334000*0.0144 = 4809.6 J,

where Cmi is specific heat of melting ice.

E3 = Cw*(61-0)*0.0144 = 4218*(61-0)*0.0144 = 3705.0912 J,

where Cw = 4218 J/kg*K is the specific heat of a clean water.

So, we need E = E1 + E2 + E3 = 2430.72 + 4809.6 + 3705.0912 = 10945.4112 J = 1.09454112 kJ to convert 14.4 grams of ice at -80.0 degrees Celsius to water at 61 degrees Celsius.

where Ci = 1800 J/kg*K is the average specific heat of an ice (from -80 to 0 degree).

E2 = Cmi*0.0144 = 334000*0.0144 = 4809.6 J,

where Cmi is specific heat of melting ice.

E3 = Cw*(61-0)*0.0144 = 4218*(61-0)*0.0144 = 3705.0912 J,

where Cw = 4218 J/kg*K is the specific heat of a clean water.

So, we need E = E1 + E2 + E3 = 2430.72 + 4809.6 + 3705.0912 = 10945.4112 J = 1.09454112 kJ to convert 14.4 grams of ice at -80.0 degrees Celsius to water at 61 degrees Celsius.

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