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.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1