Amass of 32.0 g of h2o(g) at 373 k is flowed into 295 g of h2o(l) at 310. k and 1 atm. calculate the final tem- perature of the system once equilibrium has been reached. assume that cp, m for h2o is constant at its values for 298 k throughout the temperature range of interest. describe the state of the system.

313 K;

Liquid state.

Explanation:

When the systems are put in contact they intend to achieve thermal equilibrium, it means that they will change heat until they both have the same final temperature (T). If the system is isolated, by the energy equilibrium, the total heat change must be 0. Thus, the water vapor (hotter) loses heat to the liquid water.

Qv + Ql = 0

The heat is the heat capacity at constant pressure (Cp) multiplied by the mass and by the change in temperature (final - initial). Looking at a thermodynamics table, the values of Cp at 298K for water vapor and liquid are:

Cpv = 0.480 cal/g.K

Cpw = 1.000 cal/g.K

Cpv*mv*(T - 373) + Cpw*mw*(T - 310) = 0

0.480*32.0*(T-373) = -1.000*295*(T-310)

15.36*(T - 373) = -295*(T-310)

15.36T - 5729.28 = -295T + 91450

310.32T = 97179.28

T = 313 K

Knowing that the fusion temperature of the water is 273 K and the boiling point is at 373 K, the system at 313 K is ar the liquid state.

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question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

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(jacemorris04)