When running at a constant temperature, one practical engine goes through 2400

cycles every minute. In one complete cycle of this engine, 114 J of energy has to be

removed by a coolant so that the engine runs at a constant temperature. The

temperature of the coolant rises by 18 °C as it passes through the engine.

Calculate the volume of the coolant that flows through the engine in one second.

specific heat capacity of coolant = 3.8 × 10^3 J kg–1 K–1

density of coolant = 1.1 × 103 kg m–3

Can someone please help me with this question I can't seem to get the same answer as the one in the mark scheme

-so first I found Q= (2400/60)x 114 = 4560J

-then I used Q=mcT with T=18+273 to find the mass

-then I used density=m/v to rearrange for volume and I got 3.8x10^-6 and the answer is meant to be = 6.06 × 10^−5

cycles every minute. In one complete cycle of this engine, 114 J of energy has to be

removed by a coolant so that the engine runs at a constant temperature. The

temperature of the coolant rises by 18 °C as it passes through the engine.

Calculate the volume of the coolant that flows through the engine in one second.

specific heat capacity of coolant = 3.8 × 10^3 J kg–1 K–1

density of coolant = 1.1 × 103 kg m–3

Can someone please help me with this question I can't seem to get the same answer as the one in the mark scheme

-so first I found Q= (2400/60)x 114 = 4560J

-then I used Q=mcT with T=18+273 to find the mass

-then I used density=m/v to rearrange for volume and I got 3.8x10^-6 and the answer is meant to be = 6.06 × 10^−5

Original post by 1234kelly

When running at a constant temperature, one practical engine goes through 2400

cycles every minute. In one complete cycle of this engine, 114 J of energy has to be

removed by a coolant so that the engine runs at a constant temperature. The

temperature of the coolant rises by 18 °C as it passes through the engine.

Calculate the volume of the coolant that flows through the engine in one second.

specific heat capacity of coolant = 3.8 × 10^3 J kg–1 K–1

density of coolant = 1.1 × 103 kg m–3

Can someone please help me with this question I can't seem to get the same answer as the one in the mark scheme

-so first I found Q= (2400/60)x 114 = 4560J

-then I used Q=mcT with T=18+273 to find the mass

-then I used density=m/v to rearrange for volume and I got 3.8x10^-6 and the answer is meant to be = 6.06 × 10^−5

cycles every minute. In one complete cycle of this engine, 114 J of energy has to be

removed by a coolant so that the engine runs at a constant temperature. The

temperature of the coolant rises by 18 °C as it passes through the engine.

Calculate the volume of the coolant that flows through the engine in one second.

specific heat capacity of coolant = 3.8 × 10^3 J kg–1 K–1

density of coolant = 1.1 × 103 kg m–3

Can someone please help me with this question I can't seem to get the same answer as the one in the mark scheme

-so first I found Q= (2400/60)x 114 = 4560J

-then I used Q=mcT with T=18+273 to find the mass

-then I used density=m/v to rearrange for volume and I got 3.8x10^-6 and the answer is meant to be = 6.06 × 10^−5

Your mistake is using the equation Q = mcΔT incorrectly. It is the change in temperature NOT the absolute temperature that we are using in this equation.

I believe you are doing A level. You should be more careful in writing the units that you have at the end of every single calculation.

Q = (2400/60) × 114 = 4560J/s

Original post by Eimmanuel

Your mistake is using the equation Q = mcΔT incorrectly. It is the change in temperature NOT the absolute temperature that we are using in this equation.

I believe you are doing A level. You should be more careful in writing the units that you have at the end of every single calculation.

Q = (2400/60) × 114 = 4560J/s

I believe you are doing A level. You should be more careful in writing the units that you have at the end of every single calculation.

Q = (2400/60) × 114 = 4560J/s

(edited 1 month ago)

Original post by 1234kelly

Thanks for the reply! But quick question I thought I did use the change in temperature, the coolant rises by 18c ? And could you explain a bit more about what the units of Q have to do with where I went wrong please I don’t really understand what I am meant to do with the fact Q is in J/s ?

No more no less.

Not sure why you add 273 to 18.

When you compute Q using (2400/60) × 114, what is the units for (2400/60)?

Original post by Eimmanuel

When the temperature of coolant rises by 18°C, the change in temperature is just 18°C.

No more no less.

Not sure why you add 273 to 18.

When you compute Q using (2400/60) × 114, what is the units for (2400/60)?

No more no less.

Not sure why you add 273 to 18.

When you compute Q using (2400/60) × 114, what is the units for (2400/60)?

Original post by 1234kelly

-I added the 273 to convert it into Kelvin

-I got that Q= (2400/60)x 114 but then what do I do then to get the volume if my initial method was wrong there is no other way to workout the mass but use Q=mcT no ?

-I got that Q= (2400/60)x 114 but then what do I do then to get the volume if my initial method was wrong there is no other way to workout the mass but use Q=mcT no ?

1 °C = 1 K

This is why in calculating Q = m*c*Delta T, we don't need to comvert °C to K for temperature.

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