Question

A total of 619 cal of heat is added to 5.00 g of ice at −20.0 °C. What is the final temperature of the water? Specific heat of ?2?(?) 2.087 J/(g⋅°C) Specific heat of ?2?(?) 4.184 J/(g⋅°C) Heat of fusion for ?2? 333.6 J/g

Answer 1

The following flowchart shows the what happens to ice when 619 cal of heat is supplied. For the rest of the calculation we will convert this into Joule.

Total heat supplied=619 cal=619*4.184J =2589.89J

Ice(-20^oC)----->Ice(0^oC)----->Water(0^oC)----->Water(T_f)

Heat required for converting ice at -20^oC to ice at 0^oC = m*C₁*O₁=5*2.087*20=208.7J

Where m= mass of ice= 5.00g , C₁= specific heat of ice=2.087J/g^oC & O₁= change in temperature=0-(-20)=20^oC

Heating required for melting ice=m*T_m=5*333.6=1668J

Where m= mass of ice= 5.00g , Tm= Heat of fusion of ice= 333.6J/g

Therefore total heat for converting ice at at -20^oC to water at 0^oC=1668+208.7=1876.7J

Amount of heat left=2589.89-1876.7=713.19 J

This is the amount of heat required to raise the temperature of water at 0^oC

Amount of heat required to raise the temperature of water at 0^oC= m* C₂ *O₂= 5*4.184 *O₂

Where m= mass of ice= 5.00g , C₂= specific heat of water=4.134J/g^oC & O₂= change in temperature=T_f- 0=T_f

[T_f is the final temperature]

Equating these two values,

713.19= 5*4.184*T_f

T_f = 713.19/(5*4.184)= 34.09^oC