You are a professional frontier re-enactor who specializes in making lead shot.

Question

You are a professional frontier re-enactor who specializes in making lead shot. To do this you put 2 kg of ice into 2 kg of water, both at 0 C, into an insulated bucket. To make the shot, you carefully drop 1 g droplets of molten lead which is exactly at the melting temperature of 327.5 C into the bucket of ice-water. The molten lead solidifies into perfect spheres and fall to the bottom of the bucket.

c-lead = .13kJ/kg*C

c-H2O = 2.1kJ/kg*C

Lf-lead = 22.4kJ/kg

Lf-H2O = 334kJ/kg

How much energy is required to cool a single droplet of molten lead all the way down to 0 C? (In Joules)

What is the maximum number of lead shot balls can you make without changing the ice-water temperature from 0C?

Thanks in advance!

Explanation / Answer

Part A)

Q = mc(delta T)

Q = (.001)(130)(327.5)

Q = 42.6 J

Part B)

To warm up the solution, first the ice would have to be melted

Q = mH = (2)(334000) = 668000 J

Then the total amount of water would have to heat up by 1 degree

Q = mc(delta T)

Q = 4(4190)(1) = 16760 J

Total Enegry = 16760 + 668000 = 684760 J

Divide that by 42.6 and get 684760/42.6 = 16074.2 shot

That means you can make 16.074 shot without changing the temp.

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