A 30.0-g ice cube at its melting point is dropped into an aluminum calorimeter of mass 100.0 g in equilibrium at 24.0 °C with 300.0 g of an unknown liquid. The final temperature is 4.0 °C . What is the heat capacity of the liquid?

Cu = 1453.72J/Kg°C

The heat capacity of the liquid is 1453.72J/Kg°C

Explanation:

At equilibrium, assuming no heat loss to the surrounding we can say that;

Heat gained by ice + heat gained by cold water = heat loss by hot unknown liquid (24°C) + heat loss by aluminium calorimeter.

Given;

Mass of ice = mass of cold water = mc = 30g = 0.03kg

Mass of hot unknown liquid mh= 300g = 0.3kg

Mass of aluminium calorimeter ma= 100g = 0.1 kg

change in temperature cold ∆Tc = (4-0) = 4°C

Change in temperature hot ∆Th = 24-4 = 20°C

Specific heat capacity of water Cw= 4186J/Kg°C

Specific heat capacity of aluminium Ca = 900J/kg°C

Specific heat capacity of unknown liquid Cu =?

Heat of condensation of ice Li = 334000J/Kg

So, the statement above can be written as.

mcLi + mcCw∆Tc = maCa∆Th + mhCu∆Th

Making Cu the subject of formula, we have;

Cu = [mcLi + mcCw∆Tc - maCa∆Th]/mh∆Th

Substituting the values we have;

Cu = (0.03×334000 + 0.03×4186×4 - 0.1×900×20)/(0.3×20)

Cu = 1453.72J/Kg°C

the heat capacity of the liquid is 1453.72J/Kg°C

howdy howdy!

i'd say it would be galileo!

and 1and 1