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.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'