200 g of ice at 0°c is dropped into a calorimeter of negligible heat capacity containing 350 g of water at 20°c. what is the temperature of the system when it reaches equilibrium? the latent heat of fusion of water is 80 cal/g.

The temperature of the system when it reaches equilibrium is 0°C.

Explanation:

Heat gained by ice at 0 °C to convert into 0 °C water is:

Mass of ice at 0 °C = 200g

Enthalpy of fusion is defined as the the heat gain when 1 g of the substance at its freezing point melts to liquid under standard pressure.

Thus heat gained by water = 80 calg⁻¹×200 g = 16000 kcal

Heat lost by water at 20 °C to convert to 0 °C:

The relation between heat released/gained in a system and the temperature is:

where,  

Q=mcΔT

Q is the amount of heat absorbed or released

m is the mass

ΔT is the change in temperature

c is called the specific heat.  

Specific heat is defined as heat gained by 1 unit mass of any sample to raise the temperature by 1 °Celsius.

Thus, from the question:

Mass of water =350 g

ΔT = (20 -0)°C = 20°C

Specific heat of water = 1 cal/g°C

Thus, Heat lost by water :

Q = 350 g×1 cal/g°C×20°C = 7000 cal

Thus, Heat lost < Heat gained, the system is yet at equilibrium. Not all the ice melt in the system.

Thus, The temperature of the system when it reaches equilibrium is 0°C.

i believe the answer is three water molecules

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.