Answer:
Total heat energy input needed is 6116kJ
Step-by-step explanation:
This problem bothers heat capacity
Assuming that the steel container and the liquid have same temperatures
Given data
Initial temperature of liquid and steel container θ1=20°C
Final temperature of liquid and steel containers θ2= 100°C
Mass of container M1 =15kg
Mass of liquid M2= 17kg
Specific heat capacity of steel container cst450 J/kg.C
Specific heat capacity of
Liquid cli = 4100 J/kg.C
We know that the heat applied is given as
Q= mcΔθ
Total heat applied = Ql + Qs
Where Ql= heat applied to liquid
Qs= heat applied to steel
Substituting our given data we have
Q= 15*450(100-20)+17*4100(100-20)
Q= 6750(80)+69700(80)
Q=540000+5576000
Q= 6116000
Q= 6116kJ