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I'm in desperate need of some help!
I'm sitting an exam when I go back on Thermodynamics (temperature change graphs - read the change off the graph to do calculations on it) and I need some recap on q=mc^t and what I put into it. Please help!
I'm sitting an exam when I go back on Thermodynamics (temperature change graphs - read the change off the graph to do calculations on it) and I need some recap on q=mc^t and what I put into it. Please help!
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#2
What's the problem? I'll give a quick overview!
q = mc (delta T)
q = heat energy absorbed/evolved (J)
m = mass of substance (g)
c = heat capacity (J g ^-1 K^-1)
delta T = temperature change
Example question: 100 g of water is heated from 25°C to 75°C on an electric heater. How much energy is transferred in this process? Assume the heat capacity of water is 4.18 J g^-1 K^-1.
Easy!
m = 100
c = 4.18
delta T = 50
q = mc (delta T)
= (100)(4.18)(50)
= 20 900 J = +20.9 kJ (DON'T FORGET THE POSITIVE SIGN - Endothermic process, heat is absorbed!)
A beaker of water is cooled from 65°C to 35°C. Calorimeter suggests that 37.62 kJ of heat energy is given off in this process. Assuming that no energy is lost (the energy measured is 100% of the energy given off), calculate the mass of the water in the beaker.
Easy again - simply rearrange for mass!
If q = mc deltaT, then m = q / c deltaT
q = -37 620
c = 4.18 (same substance!)
delta T = -30
m = (- 37 620) / (4.18)(-30)
= 300 g of water!
q = mc (delta T)
q = heat energy absorbed/evolved (J)
m = mass of substance (g)
c = heat capacity (J g ^-1 K^-1)
delta T = temperature change
Example question: 100 g of water is heated from 25°C to 75°C on an electric heater. How much energy is transferred in this process? Assume the heat capacity of water is 4.18 J g^-1 K^-1.
Easy!
m = 100
c = 4.18
delta T = 50
q = mc (delta T)
= (100)(4.18)(50)
= 20 900 J = +20.9 kJ (DON'T FORGET THE POSITIVE SIGN - Endothermic process, heat is absorbed!)
A beaker of water is cooled from 65°C to 35°C. Calorimeter suggests that 37.62 kJ of heat energy is given off in this process. Assuming that no energy is lost (the energy measured is 100% of the energy given off), calculate the mass of the water in the beaker.
Easy again - simply rearrange for mass!
If q = mc deltaT, then m = q / c deltaT
q = -37 620
c = 4.18 (same substance!)
delta T = -30
m = (- 37 620) / (4.18)(-30)
= 300 g of water!
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(Original post by mobius323)
What's the problem? I'll give a quick overview!
q = mc (delta T)
q = heat energy absorbed/evolved (J)
m = mass of substance (g)
c = heat capacity (J g ^-1 K^-1)
delta T = temperature change
Example question: 100 g of water is heated from 25°C to 75°C on an electric heater. How much energy is transferred in this process? Assume the heat capacity of water is 4.18 J g^-1 K^-1.
Easy!
m = 100
c = 4.18
delta T = 50
q = mc (delta T)
= (100)(4.18)(50)
= 20 900 J = +20.9 kJ (DON'T FORGET THE POSITIVE SIGN - Endothermic process, heat is absorbed!)
A beaker of water is cooled from 65°C to 35°C. Calorimeter suggests that 37.62 kJ of heat energy is given off in this process. Assuming that no energy is lost (the energy measured is 100% of the energy given off), calculate the mass of the water in the beaker.
Easy again - simply rearrange for mass!
If q = mc deltaT, then m = q / c deltaT
q = -37 620
c = 4.18 (same substance!)
delta T = -30
m = (- 37 620) / (4.18)(-30)
= 300 g of water!
What's the problem? I'll give a quick overview!
q = mc (delta T)
q = heat energy absorbed/evolved (J)
m = mass of substance (g)
c = heat capacity (J g ^-1 K^-1)
delta T = temperature change
Example question: 100 g of water is heated from 25°C to 75°C on an electric heater. How much energy is transferred in this process? Assume the heat capacity of water is 4.18 J g^-1 K^-1.
Easy!
m = 100
c = 4.18
delta T = 50
q = mc (delta T)
= (100)(4.18)(50)
= 20 900 J = +20.9 kJ (DON'T FORGET THE POSITIVE SIGN - Endothermic process, heat is absorbed!)
A beaker of water is cooled from 65°C to 35°C. Calorimeter suggests that 37.62 kJ of heat energy is given off in this process. Assuming that no energy is lost (the energy measured is 100% of the energy given off), calculate the mass of the water in the beaker.
Easy again - simply rearrange for mass!
If q = mc deltaT, then m = q / c deltaT
q = -37 620
c = 4.18 (same substance!)
delta T = -30
m = (- 37 620) / (4.18)(-30)
= 300 g of water!
Quick question. When it's a powder being added to a known substance of acid, what's the weight? Is it the acid weight, the powder weight, or both added together?
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#4
(Original post by WishfulDesire)
Thanks!
Quick question. When it's a powder being added to a known substance of acid, what's the weight? Is it the acid weight, the powder weight, or both added together?
Thanks!
Quick question. When it's a powder being added to a known substance of acid, what's the weight? Is it the acid weight, the powder weight, or both added together?
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#5
(Original post by mobius323)
What's the problem? I'll give a quick overview!
q = mc (delta T)
q = heat energy absorbed/evolved (J)
m = mass of substance (g)
c = heat capacity (J g ^-1 K^-1)
delta T = temperature change
Example question: 100 g of water is heated from 25°C to 75°C on an electric heater. How much energy is transferred in this process? Assume the heat capacity of water is 4.18 J g^-1 K^-1.
Easy!
m = 100
c = 4.18
delta T = 50
q = mc (delta T)
= (100)(4.18)(50)
= 20 900 J = +20.9 kJ (DON'T FORGET THE POSITIVE SIGN - Endothermic process, heat is absorbed!)
A beaker of water is cooled from 65°C to 35°C. Calorimeter suggests that 37.62 kJ of heat energy is given off in this process. Assuming that no energy is lost (the energy measured is 100% of the energy given off), calculate the mass of the water in the beaker.
Easy again - simply rearrange for mass!
If q = mc deltaT, then m = q / c deltaT
q = -37 620
c = 4.18 (same substance!)
delta T = -30
m = (- 37 620) / (4.18)(-30)
= 300 g of water!
What's the problem? I'll give a quick overview!
q = mc (delta T)
q = heat energy absorbed/evolved (J)
m = mass of substance (g)
c = heat capacity (J g ^-1 K^-1)
delta T = temperature change
Example question: 100 g of water is heated from 25°C to 75°C on an electric heater. How much energy is transferred in this process? Assume the heat capacity of water is 4.18 J g^-1 K^-1.
Easy!
m = 100
c = 4.18
delta T = 50
q = mc (delta T)
= (100)(4.18)(50)
= 20 900 J = +20.9 kJ (DON'T FORGET THE POSITIVE SIGN - Endothermic process, heat is absorbed!)
A beaker of water is cooled from 65°C to 35°C. Calorimeter suggests that 37.62 kJ of heat energy is given off in this process. Assuming that no energy is lost (the energy measured is 100% of the energy given off), calculate the mass of the water in the beaker.
Easy again - simply rearrange for mass!
If q = mc deltaT, then m = q / c deltaT
q = -37 620
c = 4.18 (same substance!)
delta T = -30
m = (- 37 620) / (4.18)(-30)
= 300 g of water!
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(Original post by Lallana 11)
its both added together
its both added together
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#8
Umm not sure what you mean but the reaction equation doesn't make a difference when you use the weight in the q=mc(delta)T
You just add the mass of the substance to the volume of acid that they've used , eg 5g of solid in 50 cm^3 of HCl
Means you just add 5 and 50 =55g
You just add the mass of the substance to the volume of acid that they've used , eg 5g of solid in 50 cm^3 of HCl
Means you just add 5 and 50 =55g
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