# Homework Help: Latent heat of Fusion Lab

1. Nov 7, 2007

### srose9625

A student takes 0.45kg of ice. The ice is initally at -23'C. She heats the sample in an oven until the temp is 134'C.
a). What is the NRG needed to heat the ice to its melting point?
b). What is the NRG needed to change the solid ice into liquid water?
c). What is the NRG needed to heat the liquid water to its boiling point?
d). What is the NRG needed to change the liquid water to steam?
e). What is the NRG needed to heat the steam to 134'C?
f). Draw a graph with the temp of the sample in Kelvin on y axis, and NRG absorbed in Joules by the sample on the x axis.

Equations of use:
Q=mc(change in T)
Q=mL

a) (.45kg) (4.2x10^3 J/kg) (-23'C)=-4.3x10^6 J
b) (.45kg) (3.3x10^5 J/kg) = 1.5x10^5 J
c) (.45kg) (3.3x10^5 J/kg) (134'C) = 1.2x10^7 J
d) (.45kg) (2.3x10^6 J/kg) = 1.0x10^6 J
e) (.45kg) (2.3x10^6 J/kg) (134'C) = 1.4x10^8 J
f) I have ice drawn in a straight/horizontal line, then water drawn in a linear line followed by a straight/horizontal line, then steam in a linear line, then boiling in a straight/horizontal line.

Last edited: Nov 7, 2007
2. Nov 7, 2007

### Chi Meson

a) b) and d) are correct but, why is a) a negative answer? The change in temperature is +, going from -23 to zero.

and c). Do you know the temperature at which liquid water will boil?

And please don't say "NRG." Use "Q" which stands for "heat."

3. Nov 7, 2007

### srose9625

a). (.45kg) (4.2x10^3 J/kg) (23'C) = 4.3x10^6 J
Is that still right?

c). (.45kg) (3.3x10^5 J/kg) (100'C) = 1.2x10^7 J
I am guessing that water boils at 100'C. Would that be the right answer?

e). (.45kg) (3.3x10^5 J/kg) (134'C) = 1.2x10^7 J
OR
(.45kg) (2.3x10^6 J/kg) (134'C) = 1.4x10^8 J

f). I have ice drawn in a straight/horizontal line, then water drawn in a linear line followed by a straight/horizontal line, then steam in a linear line, then boiling in a straight/horizontal line.