1 A student drops a 035kg piece of steel at 42 C into a cont

1. A student drops a 0.35-kg piece of steel at 42 C into a container of water at 22 C. The student also drops a 0.53-kg chunk of lead into the same container at the same time. The temperature of the water remains the same. What was the temperature of the lead?

2. A 370-mL spherical flask contains 0.065 mol of an ideal gas at a temperature of 283 K. What is the average force exerted on the walls of the flask by a single molecule (in Newtons)?

3. A 4.9-kg block of ice at -1.5 C slides on a horizontal surface with a coefficient of kinetic friction equal to 0.069. The initial speed of the block is 7.2 m/s and its final speed is 5.5 m/s. Assuming that all the energy dissipated by kinetic friction goes into melting a small mass m of the ice, and that the rest of the ice block remains at -1.5 C, determine the value of m (in kg).

Solution

Q.1. Finally all three comes at temperature 22 degree celcius in equilibrium.

So heat lost by steel = m x c x (42-22) = 0.35 x 0.49 x 20 = 3.43 KJ

Because Cp od steel =  0.49 KJ/(Kg-K)

This heat is gained by Lead.

So, m x c x (22-T) = 3.43

or 0.53 x 0.13 x (22-T) = 3.43

or T= The temperature of the lead when dropped = -27.78 degree celcius