The basic equation used to determine the amount of energy needed to change the temperature of a given mass of a substance is:
#q=mxxcxxDeltaT#,
where:
#q# is energy, #m# is mass, #c# is specific heat, and #DeltaT# is change in temperature. #DeltaT=(T_"final" - T_"initial")#.
There are four steps required to answer this question. First, the energy required to raise the temperature of the ice from #-30.00^@"C"# to #0.00^@"C"# must be determined. Then the energy required to melt the ice must be determined. Then the energy required to raise the temperature from #0.00^@"C"# to #85.00^@"C"# must be determined. Once you have all three values for energy, add them together to get the total energy.
1. Energy required to change the temperature from #-30.00^@"C"# to #0^@"K"#. In this equation we will use the specific heat of solid water (ice).
Known
#m="36.00 g"#
#c_"ice"=(2.108"J")/("g"*""^@"C")#
#T_"i"=-30.00^@"C"#
#T_"f"=0.00^@"C"#
#DeltaT=0.00^@"C"-(- 30.00^@"C")="30.00"^@"C"#
Solution
Plug the known values into the equation and solve for #q#.
#q = (36.00color(red)cancel(color(black)("g")))xx((2.108"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))xx(30.00^@color(red)cancel(color(black)("C")))="2277 J"# (rounded to four significant figures)
2. Energy required to melt the ice.
Equation:
#q="m"xxH_"f"#,
where:
#q# is energy, #H_"f"# is the heat of fusion, and #m# is mass.
The temperature is not needed because the temperature doesn't change during a phase change.
Known
#H_"f"="334 J/g"#
#m="36.00 g"#
Solution
Plug in the known values and solve for #q#.
#q=36.00color(red)cancel(color(black)("g"))xx(334 "J")/color(red)cancel(color(black)("g"))="12020 J"# (rounded to four significant figures)
3. Energy required to change the temperature from #"0.00"^@"C"# to #"85"^@"C"#. In this equation we will use the specific heat for liquid water.
Equation:
#q=mcDeltaT#
Known
#c_"H2O"=(4.184 "J")/("g"*""^@"C")"#
#m="36.00 g"#
#T_"i"="0.00"^@"C"#
#T_"f"="85.00"^@"C"#
#DeltaT="85.00"^@"C"-"0.00"^@"C"="85.00"^@"C"#
Solution
Plug in the known values and solve for #q#.
#q=36.00color(red)cancel(color(black)("g"))xx(4.184"J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx85.00^@color(red)cancel(color(black)("C"))= "1280. J"# (rounded to four significant figures)
4. Add the energies from each of the three equations. This will give you the total amount of energy needed to raise the temperature of a #"36.00 g"# sample of ice from #-30.00^@"C"# to #"85.00"^@"C"#.
#"Total energy"="2277 J" + "12020 J" + "1280 J"="15577 J"# (rounded to a whole number)
