In: Physics
2) Given the rising price of commodities, I have found it profitable to melt my copper pennies for scrap. I’ve saved about 150 kg of them, and I get down to business. The melting point of copper is 1085 degrees Celsius. However, my stove runs out of fuel after putting out only 28,450,000 J of heat.
A. Assuming the pennies started at 20 degrees Celsius, what percentage has melted and how much is still solid?
B. 1 kilogram of propane delivers 40,000,000 J of heat when burned. How much more propane do I need?
Given data
M=150kg
T2 =20
T1=1085
A)
Copper is not given from question i have considered a value from table as value of 385 J/KgoC
We know that the formula for Q is
Q = ms( T2 - T1) |
Q = 150(385)(1085-20)
Q=57750(1065)
Q = 6.15 X 107 J
We know that the
Percentage of copper melted = (amount of heat supplied by the stove before it runs out of fuel)/(Amount of heat required to melt 150 Kg of pennies) |
The amount of heat supplied by the stove before it runs out of fuel is 28,450,000 J
Percentage of copper melted = (28,450,000 / 6.15 X 107) X 100
2845/6.15 X 102
Percentage of copper melted = 46.26 %
Percentage of copper that is still solid = 100 -46.26 = 53.74 % |
B)
Given 1 kilogram of propane delivers 40,000,000J of heat
So, Amount of heat required to melt 150 Kg of pennies = 6.15 X 107 J
Already supplied heat before it runs out of fuel is 28,450,000 J
By giving additional heat required= 6.15 X 107 - 28,450,000
= 61500000-28450000 =33050000/ 107
= 3.305 X 107 J
The additional amount of heat required = 3.305 X 107 J
Mass of Propane required = 3.305 X 107 J / 40,000,000J
Mass of Propane required = 0.83 Kg
0.83 Kg of more propane is needed
Mass of Propane required =1.54 Kg |