Question

A large insulated bottle containing 703 mL of water initially at 21.5 degrees C is cooled by adding ice at 0 degrees C. Assume the specific heat of liquid water is constant at 4.18 kJkg K

(a) If 82 grams of ice is added to the bottle, what will the final temperature be? (Degrees C)

(b) How much ice must be added if the desired temperatue is 5.1 degrees C? (grams)

Answer 1

Now here the system is insulated, hence we have an adiabatic system. meaning there is no transfer of heat between system and its surrounding.

Volume of water = 703 ml = 0.703 L = 0.703*10^-3 m³​

Density of water = 1000 Kg/m³

​Mass of water ( M_Water) = Density of water * Volume of water = 1000 * 0.703*10^-3 = 0.703 Kg

Cp of water = 4.18 KJ/Kg K

Initial temperature of water ( Ti) = 21.5 ^oC

Heat of fusion of ice (Hf) = 334 J = 3334 KJ/Kg

we will make use of general enthalpy balance

Heat given by ice = heat absorbed by water

Let initial temperature fo ice be "To = 0 ^oC"

Let the final temperature be T ^oC

Let mass of ice which melts be M_ice

a) Make enthalpy balance

M_ice ( Hf + Cp*dT) = M_Water *Cp * dT

0.082*(334 + 4.18*(T-0)) = 0.703*4.18*(21.5-T)

solving the above equation for T we get, T =10.91 ^oC

b) here again we will apply the same enthalpy balance, except here we are given the final temperature of the mixture T and we need to find the mass of ice (M_ice) to attain that temperature

M_ice ( Hf + Cp*dT) = M_Water *Cp * dT

M_ice*(334 + 4.18*(5.1-0)) = 0.703*4.18*(21.5-5.1)

Solving the above equation for M_ice we get, M_ice = 0.1356 Kg =135.6 grams