If 50 g of ice at −5

Question

◦C is mixed into 300 g of water at 60◦C in a completely insulated container,
what is the final equilibrium temperature of the water? Why did we insist on the container
being completely insulated?

Answer 1

For this kind of set up, we have summation of all energy transferred equal to zero. Note that for ice, Q is different from each process: (1) from T=-5 C to T=0 ; (2) phase change of ice to water at constant T ; (3) from T=0 to final T.
For Ice:
(1) from T=-5 C to T=0. We use the formula:
Q = mcΔT
where
m = mass (g)
c = specific heat capacity (J/g-K)
T = temperature (K)

Q = ( 50 g )( 2.01 J/g-K )( 0 - (-5) K)
Q = 502.5 J

(2) phase change of ice to water at constant T
Q = mΔHf
where Hf is the latent heat of fusion
Q = ( 50 g )( 333.33 J/g )
Q = 16666.67 J

(3) from T=0 to final T. At this point, ice is now water:
Q = mcΔT
Q = ( 50 g )( 4.184 J/g-K )(T - 0)
Q = 209.2T

For hot water initially at 60 C:
Q = mcΔT
Q = ( 300 g )( 4.184 J/g-K )(T - 60)
Q = 1255.2(T - 60)

Combining all Q's:
502.5 + 16666.67 + 209.2T + 1255.2(T - 60) = 0

Now solve for T. Units in degree Celsius.
The system must be insulated so that no energy could escape into the surroundings, therefore heat absorbed and heat lost are only happening with ice and hot water.
By the way the values for latent heat and heat capacities can be found on google or textbooks.
hope this helps~ `u`