Question

In: Physics

2) Given the rising price of commodities, I have found it profitable to melt my copper...

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?

Solutions

Expert Solution

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


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