Asked by Tracy
Calculate the quantity of heat to convert 25.0g of ice at -10.0 degrees celcius to steam at 110.0 degrees celcius?
Answers
Answered by
DrBob222
Note the correct spelling of celsius.
q1 = heat to move ice from -10 to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = 25 x ?? x [0-(-10)]
q2 = heat to melt the ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.
q3 = heat to move liquid water at zero C to 100 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial).
q4 = heat to convert water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization.
q5 = heat to move steam at 100 C to 110 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial).
Total Q = q1+q2+q3+q4+q5.
q1 = heat to move ice from -10 to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = 25 x ?? x [0-(-10)]
q2 = heat to melt the ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.
q3 = heat to move liquid water at zero C to 100 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial).
q4 = heat to convert water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization.
q5 = heat to move steam at 100 C to 110 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial).
Total Q = q1+q2+q3+q4+q5.
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