(a) In a question like this, you will be given that the Specific Heat Capacity of water is 4200Jkg-1K-1. However, commiting this to memory will be helpful.
Start with the equation:
E=mC(T1-T0)
Find which variables you have and convert to Standard Units:
E=What we want.
m=175g=0.175kg
C=4200Jkg-1K-1
T0=85oC=358oK
T1=74oC=347oK
Substitute in and E=0.1754200(347-358)=-8085J or =-8.085kJ *Negative indicates that the energy is LOST.
(b) Use the equation again and find which variables you know. Convert to Standard Units again.
E=+8085J The energy LOST by the water must be GAINED by the milk. (Assume none is lost to the surroundings?
m=What we want
C=3900Jkg-1K-1 (You will not be expected to know this, it will be given in the question)
(T1-T0)=(74+273)-(4.5+273)=69.5oK
Substitute in: 8085=m390069.5
Rearrange for m => m=8085/(3900*69.5)=0.0298kg (3s.f.)