A100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. when the mixture reaches equillibrium, the temperature of the mixture is 11 degrees celsius. from this information find the latent heat of fusion of water ( specific heat of water is 4186 j/kg degree celsius)

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Thus total energy needed equals

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Now by conservation of energy we have

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

the force on the thing and the distance the force moves the thing