Question

A thirsty nurse cools a 2.00 L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.249 kg and adding 0.123 kg of ice initially at -14.2 ∘C. If the soft drink and mug are initially at 20.7 ∘C, what is the final temperature of the system, assuming no heat losses? Express your answer in degrees Celsius to three significant figures.

Answer 1

we know,

C_aluminum = 900 J/(kg C)

C_water = 4186 J/(kg C)

C_ice = 2030 J/(kg C)

Lf = 3.33*10^5 J/kg

Let T is the final temperature

Apply,

heat lost by drink and mug = heat gained by ice

2.0*4186*(20.7 - T) + 0.249*900*(20.7 - T) = 0.123*2030*(14.2-0) + 0.123*3.33*10^5 + 0.123*4186*(T - 0)

2.0*4186*20.7 - 2.0*4186*T + 0.249*900*20.7 - 0.249*900*T = 0.123*2030*(14.2-0) + 0.123*3.33*10^5 + 0.123*4186*T

2.0*4186*20.7 + 0.249*900*20.7 - 0.123*2030*(14.2-0) - 0.123*3.33*10^5 = T*(0.123*4186 + 2.0*4186 + 0.249*900)

T = (2.0*4186*20.7 + 0.249*900*20.7 - 0.123*2030*(14.2-0) - 0.123*3.33*10^5)/(0.123*4186 + 2.0*4186 + 0.249*900)

= 14.645 C