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.
