A 200 g piece of ice at 0°C is placed in 500 g of water at 20°C. The system is in a container of negligible heat capacity and insulated from its surroundings. a. What is the final equilibrium temperature of the system?
b. How much of the ice melts?

Solution :

Given data :

Mass of ice, m = 200 g

Mass of water, M = 500 g

Temperature of ice = 0°C

Temperature of water = 20°C

We known ;

The latent heat of fusion of water is

The specific heat of water, C = 4.18 kJ/kg K

a). Heat required to change the ice into water is :

  J

Heat required to cool the warm water is :

From above result,

This means that total ice could not melt.

So, the temperature must be T = 0°C

b). Q' = mL

m = 0.1253 kg

m = 125.3 g

m = 125 g

Therefore, 125 g of ice had melted.

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

answer; equivalent to a one(1) horsepower motor;