Question #42dcd

1 Answer
Mar 26, 2017

Answer:

#7.80xx10^3"cal"#, which is #7.80"Cal, kcal, food calories,nutritional calories"#, whatever

Explanation:

Recall that #q=mCDeltaT# within a phase and that during melting, #q=mH_f#

Given that #C_"water"=4.18"J/g°C"#, #C_"ice"=2.09"J/g°C"#, #H_(f"water")=334"J/g"#, and the melting point of water is #0"°C"#

First we need to make the #55.00"g"# of ice melt, that is raise it from #"-"10.0"°C"# to #0"°C"# so #DeltaT=10.0"°C"#.

#q_1=(55.00"g")(2.09"J/g°C")(10.0"°C")#, therefore #q_1=1","149.5"J"~~1","150"J"#

Then we need to actually melt the ice.

#q_2=(55.00"g")(334"J/g")#, therefore #q_2=18","370"J"~~18","370"J"#

Then we need to raise the water to the desired temperature of #57.0"°C"# from the current #0"°C"# so #DeltaT=57.0"°C"#.

#q_3=(55.00"g")(4.18"J/g°C")(57.0"°C")#, therefore #q_3=13","104.3"J"~~13","100"J"#

To find the total heat, we use #q=q_1+q_2+q_3#

#q=(1","150"J")+(18","370"J")+(13","100"J")=32","620"J"~~32","600"J"#

Recall that #1"cal"=4.18"J"#, so #(1"cal")/(4.18"J")=1#

#32","600"J"xx(1"cal")/(4.18"J")=7799.043"cal"~~7.80xx10^3"cal"#

Note: I used the #~~# symbol to show rounding for significant figures.