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 0.252 kg and adding 0.122 kg of ice initially at -14.7 ∘C. If the soft drink and mug are initially at 20.3 ∘C, what is the final temperature of the system, assuming no heat losses? Express your answer in degrees Celsius .

Answer 1

Since given that system of soft drink, ice and aluminum is isolated, So total energy will remain conserved

Suppose final temperature of system is T

Energy absorbed by ice = Energy released by soft drink + mug

Q_ice = Q_water + Q_mug

Q1 + Q2 + Q3 = Q4 + Q5

Q1 = Mi*Ci*dT1 = energy required to change -14.7 C ice into 0 C ice

Q2 = Mi*Lf = energy required to change ice into water

Q3 = Mi*Cw*dT2 = energy required to change 0 C water into T C water

Q4 = Mw*Cw*dT3 = energy required to change 20.3 C water into T C water

Q5 = Ma*Ca*dT4 = energy required to change 20.3 C mug into T C mug

Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2 = Mw*Cw*dT3 + Ma*Ca*dT4

Mi = mass of ice = 0.122 kg

Mw = mass of water (Soft drink) = 2.00 L = 2.00 kg (Since density of water is 1000 kg/m^3)

Ma = mass of aluminum mug = 0.252 kg

Using given values:

0.122*2090*(0 - (-14.7)) + 0.122*3.34*10^5 + 0.122*4186*(T - 0) = 2*4186*(20.3 - T) + 0.252*900*(20.3 - T)

T = [2*4186*20.3 + 0.252*900*20.3 - 0.122*2090*14.7 - 0.122*3.34*10^5]/(0.122*4186 + 2*4186 + 0.252*900)

T = 14.277 C

Final temperature of system = 14.3 C

Let me know if you've any query.