Question

An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?

Answer 1

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Explanation:

This is the begin:

Q1 = Q which is gained from the ice to be melted

Q2 = Q which is lost from the water to melt the ice

Q1 + Q2 = 0

We are informed that the ice is at 0 ° so we have to start calculating how many J, do we need to melt it completely. If the ice had been at a lower temperature, it should be brought to 0 ° with the formula

Q = mass. specific heat. (ΔT)

and then make the change of state by the latent heat of fusion.

The heat of fusion for water at 0 °C is approximately 334 joules per gram.

So Q = Hf . mass

Q1 = 334 J/g . 8.32 g = 2778,88 J

For water we should use this:

Q = mass. specific heat. (ΔT)

Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)

Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)

(notice we have kg, so we have to convert 55 g, to kg, 0,055kg so units can be cancelled)

Q2 = 0,055kg . 4180 J/kg. K (Tfinal (The unknown) -25°)

T° should be in K for the units of Specific heat but it is the same. The difference is the same, in K either in °C

25°C = 298K

Q2 = 0,055kg . 4180 J/kg. K (Tfinal -298K)

Now the end

Q1 + Q2 = 0

334 J/g . 8.32 g + 0,055kg . 4180 J/kg. K (Tfinal -298K)

2778,88 J + 229,9 J/K (Tfinal - 298 K) = 0

2778,88 J + 229,9 J/K x Tfinal - 68510,2 J = 0

229,9 J/K x Tfinal = 68510,2 J - 2778,88 J

229,9 J/K x Tfinal = 65731,4 J

Tfinal = 65731,4 J / 229,9 K/J

Tfinal = 285,9 K

Tfinal = 285,9 K - 273K = 12,9 °C