Question #8a0c8
1 Answer
Answer:
Explanation:
Your strategy here will be to

determine how much heat is needed to heat the sample from ice at
#5.00^@"C"# to ice at#0^@"C"# by using the specific heat of ice 
determine how much heat is needed to heat the sample from ice at
#0^@"C"# to liquid at#0^@"C"# by using the heat of fusion for water 
determine how much heat is needed to heat the sample from liquid at
#0^@"C"# to liquid at#25^@"C"# by using the specific heat of water
You will need to know

the specific heat of ice
#> 2.09"J"/("g" ""^@"C")# 
the specific heat of water
#> 4.18 "J"/("g" ""^@"C")# 
the heat of fusion for water
#> DeltaH_f = 334"J"/"g"#
So, you need to make sure that you include the phase change underwent by water when going from solid at its melting point to liquid at its melting point.
As you know, the specific heat of a substance tells you how much heat is needed to increase the temperature of
The equation that relates heat absorbed / lost and change in temperature looks like this
#color(blue)(q = m * c * DeltaT)" "# , where
So, how much heat would be needed to convert
#q_1 = 50.0 color(red)(cancel(color(black)("g"))) * 2.09 "J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [0  (5.00)]color(red)(cancel(color(black)(""^@"C")))#
#q_1 = "522.5 J"#
Now, phase changes always take place at constant temperature. The transition from solid ice at
#color(blue)(q = m * DeltaH_f)" "# , where
In your case, the heat needed for this phase change will bee equal to
#q_2 = 50.0 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g"))) = "16,700 J"#
Finally, the heat needed to heat the sample from liquid at
#q_3 = 50.0 color(red)(cancel(color(black)("g"))) * 4.18 "J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25  0)color(red)(cancel(color(black)(""^@"C")))#
#q_3 = "5,225 J"#
The total heat needed to go from ice at
#q_"total" = q_1 + q_2 + q_3#
#q_"total" = "522.5 J" + "16,700 J" + "5,225 J"#
#q_"total" = "22,447.5 J"#
I'll leave the answer expressed in kilojoules and rounded to three sig figs, despite the fact that you only have two sig figs for the final temperature of the sample
#q_"total" = color(green)("22.4 kJ")#