A mass mw = 100 g of liquid water sits in an insulated cup at temperature Tw = 1

Question

A mass mw = 100 g of liquid water sits in an insulated cup at temperature Tw = 10degreeC. Ice nt temperature Ti = -10degreeC is added, and the mixture comes to thermal equilibrium when exactly half of the ice has melted. Find the mass of ice added, tm. Solve this for the general case of any substance where the solid form of the substance is added to the liquid form. Follow the steps below: Specify that Ti = -Ts = T. where T is the number of degrees Celsius away from the substance's melting temperature. Also let alpha Ms be the mass of solid which melts during heat transfer (this means (1 - n)Ms is left unbelted), with 0

Explanation / Answer

a) if x grams of ice is added, the total weight is 100 + x. the equilibrium temperature has to be 0 degrees since that is the only temperature at which water and ice stay in equilibrium.

specific heat capacity of ice is 2.108 Kj/Kg-K;

specific heat capacity of water is 4.187 Kj/Kg-K;

latent heat of fusion for water - ice is 334 Kj/Kg

now; energy released by water while lowering its temperature from 10 to 0 degrees is : 4187 joules.

4.187 = x*10*2.108 + x/2 * 334

using this, we get x = 0.0118 kg or 11.8 gms

b) general case : cl = Ms * T * cs + alpha* Ms* Lf

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