Question

Ice (mass=26.9 g) with an initial temperature of -15.8 ^oC is added to 181.4 g of water at 27.2^oC in an isolated calorimeter. The mass of the aluminum calorimeter cup and stirrer together is 51.6 g. Using the values for specific heat and the nominal value of L_fice given in your laboratory manual, determine temperature of the system when it reaches equillibrium (T_f).

Answer 1

Suppose the final equilibrium temperature is Tf (Where Tf > 0 C)

Heat released/absorbed due to temperature change = Q = m*C*dT

m = mass, C = Specific heat capacity & dT = Tf - Ti

Now Using energy conservation:

Heat gained by ice = Heat released by water + Aluminum

Q1 + Q2 + Q3 = Q4 + Q5

mi*Ci*dT1 + mi*Lf + mi*Cw*dT3 = mw*Cw*dT4 + ma*Ca*dT5

mi = mass of ice added = 26.9 gm = 0.0269 kg

mw = mass of water = 181.4 gm = 0.1814 kg

ma = mass of calorimeter = 51.6 gm = 0.0516 kg

Ca = Specific heat capacity of aluminum cup = 900 J/kg-C

Ci = Specific heat capacity of ice = 2090 J/kg-C

Cw = Specific heat capacity of water = 4186 J/kg-C

Now using given values:

0.0269*2090*(0 - (-15.8)) + 0.0269*3.34*10^5 + 0.0269*4186*(T - 0) = 0.1814*4186*(27.2 - T) + 0.0516*900*(27.2 - T)

T = [0.1814*4186*27.2 + 0.0516*900*27.2 - 0.0269*2090*15.8 - 0.0269*3.34*10^5]/(0.0269*4186 + 0.0516*900 + 0.1814*4186)

T = 13.1 C = final temperature at equilibrium

Let me know if you've any query.