Question

A 0.0575 kg ice cube at −30.0°C is placed in 0.557 kg of 35.0°C water in a very well insulated container, like the kind we used in class. The heat of fusion of water is 3.33 x 105 J/kg, the specific heat of ice is 2090 J/(kg · K), and the specific heat of water is 4190 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system?

Answer 1

Answer:

t= 22.9ºC

Explanation:

Assuming no heat exchange outside the container, before reaching to a condition of thermal equilibrium, defined by a common final temperature, the body at a higher temperature (water at 35ºC) must give heat to the body at a lower temperature (the ice), as follows:

Qw = c*m*Δt = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t) (1)

This heat must be the same gained by the ice, which must traverse three phases before arriving at a final common temperature t:

1) The heat needed to reach in solid state to 0º, as ice:

Qi =ci*m*(0ºC-(-30ºC) = 0.0575kg*2090(J/kg.ºC)*30ºC = 3605.25 J

2) The heat needed to melt all the ice, at 0ºC:

Qf = cfw*m = 3.33*10⁵ J/kg*0.0575 kg = 19147.5 J

3) Finally, the heat gained by the mass of ice (in liquid state) in order to climb from 0º to a final common temperature t:

Qiw = c*m*Δt = 4190 (J/kg.ºC)*0.0575 kg*(t-0ºC)

So, the total heat gained by the ice  is as follows:

Qti = Qi + Qf + Qiw

⇒Qti = 3605.25 J + 19147.5 J + 240.9*t = 22753 J + 240.9*t (2)

As (1) and (2) must be equal each other, we have:

22753 J + 240.9*t = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t)

⇒ 22753 J + 240.9*t = 81684 J -2334*t

⇒ 2575*t = 81684 J- 22753 J = 58931 J

⇒ $t= \frac{58931J}{2575 J/C} = 22.9C$

⇒ t = 22.9º C