How much heat (in Joules) is needed to raise the temperature from -79oC to 57oC for 31.5g of water?
I remember my teacher using different temperatures for each equation, can someone show me what that is? thanks.
4 answers
Most likely the teacher was using Kelvin which is 273+C.
No she was doing like for the first equation for temp = (0-(-79)) and for the second equation for temp = (100-57) or something.
If she was trying to find the change in temperature it would be 79+57 =136 is the temperature change.
This is a bit more complicated that it appears. First you must realize that at -79 C, the H2O is in the form of ice. So heat must be added to raise T to zero C, heat added to melt the ice, and finally heat added to raise T from zero C to 57 C.
heat to raise T from -79 C to zero C is
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tfinal is zero C and Tinitial is -79 C.
q2 = heat to melt ice.
q2 = mass ice x heat fusion
q3 = heat to raise T from zero C t 57 C.
q3 = mass H2O x specific heat H2O x Tfinal-Tinitial) where Tfinal is 57 and Tinitial is zero C.
Then total Q = q1 + q2 + q3.
heat to raise T from -79 C to zero C is
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tfinal is zero C and Tinitial is -79 C.
q2 = heat to melt ice.
q2 = mass ice x heat fusion
q3 = heat to raise T from zero C t 57 C.
q3 = mass H2O x specific heat H2O x Tfinal-Tinitial) where Tfinal is 57 and Tinitial is zero C.
Then total Q = q1 + q2 + q3.
2 answers