# Thermodynamics A2 help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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!

0

reply

Report

#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!

0

reply

(Original post by

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!

**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!

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?

0

reply

Report

#4

(Original post by

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?

**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?

0

reply

Report

#5

**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!

0

reply

(Original post by

its both added together

**Lallana 11**)its both added together

0

reply

Report

#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

1

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top