just how a lot warmth is compelled to transform that sample of ice from solid at #-12.0^

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"C"#

In order to be able to calculate all this, you will have to recognize the values for the particular warm of ice and also that of water, and the worth of the enthalpy of fusion, #DeltaH_"fus"#, for water

#c_"ice" = 2.09"J"/("g" ""^
"C")#

#DeltaH_"fus" = 334"J"/"g"#

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

Two equations will certainly come in handy here

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - warmth absorbed / lost#m# - the mass of the sample#c# - the specific warmth of the substance#DeltaT# - the adjust in temperature, characterized as final temperature minus initial temperature

#color(blue)(q = m * DeltaH_"fus")" "#, where

#q# - warm took in / lost#m# - the mass of the sample#DeltaH_"fus"# - the enthalpy of fusion of the substance

So, to get your #"5.88-g"# sample of ice from solid at #-12.0^
"C"# to #9^
"C"#, you will have to carry out it with

#q_1 = 5.88 color(red)(cancel(color(black)("g"))) * 2.09"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^
"C")))) * <0 - (-12.0)>color(red)(cancel(color(black)(""^
"C")))#

#q_1 = "147.47 J"#

To gain you sample to undergo a solid #-># liquid phase change at its melting suggest, i.e. at #0^
"C"#, you will certainly must administer it with

#q_2 = 5.88 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g")))#

#q_2 = "1963.9 J"#

Finally, to warm the sample from #0^
"C"# to #25.0^
"C"#, you must provide it via

#q_3 = 5.88 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^
"C")))) * (25.0 - 0)color(red)(cancel(color(black)(""^
"C")))#

#q_3 = "614.46 J"#

The complete heat* forced to get your sample from ice at #-12.0^
"C"# to liquid at #25.0^
"C"# will thus be equal to

#q_"total" = q_1 + q_2 + q_3#

#q_"total" = "147.47 J" + "1963.9 J" + "614.46 J"#

#q_"total" = "2725.83 J"#

Rounded to 3 sig figs and also expressed in kilojoules, the answer will be

#q_"total" = color(green)("2.73 kJ")# 