Question

803 cal of heat is added to 5.00 g ice at –20.0 °C. What is the final temperature of the water?

SPecific heat H2O(s)= 2.087 J/(g*C)

Specific heat H2O(l)=4.184 J/(g*C)

Heat of fusion= 333.6 J/g

Answer 1

heat given = > 803 cal =3361.358 joule

The heat required to raise the temperature of ice from -20 °C to 0 °C Use the formula

q1 = mcΔT = 5*2.09*( 0--20) =209 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

q2 = m·ΔH_f = 5*334 =1670 J

Heat required to raise the temperature of 0 °C water to 100 °C water

q3 = mcΔT = 5*4.18*( 100-0) =2090

total heat =209+1670 +2090 =3969

but we provided  3361.358 so water temperature will be less than 100 °C

after water become 0-degree water remaining heat will raise its temperature

remaining heat = 3361.358 - ( 209 +1670 ) = 1482.358 J

so

1482.358 J = mcΔT  = 5*4.18* ( T-0 ) 

1482.358  =  5*4.18*  T 

T =  1482.358 /(5*4.18)

T = 70.9262201 degree answer