Rates Reinforcement Questions

1.

A solution with a density of 1033 kg/m3 is flowing through a pipe at a flow rate of 25 L.s-1. What mass of the solution passes through the pipe in 3 minutes?

First, we convert minutes into seconds:

begin mathsize 14px style 3 space m i n u t e s space cross times space 60 space s e c o n d s space equals space 180 space s e c o n d s end style

Now we multiply the rate of flow by the number of seconds:

begin mathsize 14px style 25 space cross times space 180 space equals space 4 comma 500 space L end style

1m3 is equal to 1,000 L so we divide the L by 1,000:

begin mathsize 14px style fraction numerator 4 comma 500 over denominator 1 comma 000 end fraction space equals space 4.5 space m cubed end style

And now we multiply the density by the volume to calculate the mass:

begin mathsize 14px style 4.5 space cross times space 1 comma 033 space equals space 4 comma 648.5 space k g end style

2.

A 725 mg/L sugar solution is running into a tank at the rate of 45 m3/hr. How long will it take for there to be 125 kg of sugar in the tank?

1 mg is one-thousandth of a gram, and 1kg is one thousand grams, so to convert our 725 mg into kg, we divide by one million:

begin mathsize 14px style fraction numerator 725 over denominator 1 comma 000 comma 000 end fraction space equals space 0.000725 end style

The tank is filling at 45 m3/hr, so to covert this into litres, we multiply by 1,000:

begin mathsize 14px style 45 space cross times space 1 comma 000 space equals space 45 comma 000 space L divided by h r end style

If we multiply the flow rate by the concentration, we will get the mass of sugar being delivered to the tank per hour:

begin mathsize 14px style 45 comma 000 space cross times space 0.000725 space equals space 32.625 space k g divided by h r end style

We need to know how long it will take for 125 kg to be delivered, so we divide this amount by the rate:

begin mathsize 14px style fraction numerator 125 over denominator 32.625 end fraction space equals space 3.83 space h o u r s end style

3.

We use 10 kg of ingredient with a concentration of 250 g/kg of a mystery substance to produce 50 hL of a product with 35 mg/L of the mystery substance. What is the percentage yield of the mystery substance in the product?

First, we need to calculate how much mystery substance we are adding by multiplying the quantity of ingredients by the concentration:

begin mathsize 14px style 10 space k g space cross times space 250 space g divided by k g space equals space 2 comma 500 space g end style

And now we calculate how much mystery substance we have in our product by multiplying the volume by the concentration:

begin mathsize 14px style 5 comma 000 space L space cross times space 35 space m g divided by L space equals space 175 comma 000 space m g end style

Next, we convert mg into g:

begin mathsize 14px style fraction numerator 175 comma 000 over denominator 1 comma 000 end fraction space equals space 175 space g end style

And to work out the percentage yield, we divide the amount of mystery substance in the product by the amount we added and multiply by 100:

begin mathsize 14px style fraction numerator 175 over denominator 2 comma 500 end fraction space cross times space 100 space equals space 7 percent sign end style

4.

A fluid is flowing at 3.5 m.s-1, and an operator decides to speed up the flow by adjusting the pump speed. Over 4.3 seconds, the flow increases to 5.3 m.s-1. What is the rate of acceleration caused by the pump adjustment?

First, we calculate the change in velocity:

begin mathsize 14px style 5.3 space minus space 3.5 space equals space 1.8 space m. s to the power of negative 1 end exponent end style

The acceleration is the rate of change, so we calculate it by dividing the change by the time taken for the change:

begin mathsize 14px style fraction numerator 1.8 over denominator 4.3 end fraction space equals space 0.42 space m. s to the power of negative 2 end exponent end style

 In the next section we get logarithmic! 

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