# Heat question

## Heat question

How much heat (in kJ) is required to warm 10 g of ice,initially at -10 deg. C, to steam at 110.0 deg C?

**pplantblues**- Posts : 109

Join date : 2008-11-14

## Re: Heat question

This is going off of memory from last year, so not all of the terms may be accurate.

First step: calculate the energy required to go from -10°C to 0°C (melting point).

ΔH = ΔT(C

ΔH= [-10-0](2.06)(10 g) = 206 J

Next: figure out the ΔH

ΔH = m(ΔH

ΔH = 10 g(334) = 3340 J

Then: calculate the energy required to change the temperature of now liquid water from 0°C to 100°C.

ΔH = ΔT(C

ΔH = [100-0](4.18)(10 g) = 4,180 J

Next: calculate the ΔH

ΔH= m(ΔH

ΔH= 10g(2260) = 22,600 J

Finally: calculate the energy required to heat the now gaseous water those last 10 degrees.

ΔH = ΔT(C

ΔH = [110-100](1.87)(10 g) = 187 J

Now, we take ALL of the ΔH's that we calculated and add them together:

206 J + 3340 J + 4,180 J + 22,600 J + 187 J = 30,513 J

But, the question asked for it in kJ, so dividing by 1,000 we get

First step: calculate the energy required to go from -10°C to 0°C (melting point).

ΔH = ΔT(C

_{p})(m)ΔH= [-10-0](2.06)(10 g) = 206 J

Next: figure out the ΔH

_{fus}(enthalpy of fusion)ΔH = m(ΔH

_{fus})ΔH = 10 g(334) = 3340 J

Then: calculate the energy required to change the temperature of now liquid water from 0°C to 100°C.

ΔH = ΔT(C

_{p})(m)ΔH = [100-0](4.18)(10 g) = 4,180 J

Next: calculate the ΔH

_{vap}for the liquid water turning into a gas.ΔH= m(ΔH

_{vap})ΔH= 10g(2260) = 22,600 J

Finally: calculate the energy required to heat the now gaseous water those last 10 degrees.

ΔH = ΔT(C

_{p})(m)ΔH = [110-100](1.87)(10 g) = 187 J

Now, we take ALL of the ΔH's that we calculated and add them together:

206 J + 3340 J + 4,180 J + 22,600 J + 187 J = 30,513 J

But, the question asked for it in kJ, so dividing by 1,000 we get

**30.513 kJ**. And that, my friend, is how much heat (in kJ) is required to warm 10 g of ice, initially at -10 deg. C, to steam at 110.0 deg C.**funion987**- Administrator
- Posts : 146

Join date : 2008-11-09

## Thanks for your answer

Thank you so much.

funion987 wrote:This is going off of memory from last year, so not all of the terms may be accurate.

First step: calculate the energy required to go from -10°C to 0°C (melting point).

ΔH = ΔT(C_{p})(m)

ΔH= [-10-0](2.06)(10 g) = 206 J

Next: figure out the ΔH_{fus}(enthalpy of fusion)

ΔH = m(ΔH_{fus})

ΔH = 10 g(334) = 3340 J

Then: calculate the energy required to change the temperature of now liquid water from 0°C to 100°C.

ΔH = ΔT(C_{p})(m)

ΔH = [100-0](4.18)(10 g) = 4,180 J

Next: calculate the ΔH_{vap}for the liquid water turning into a gas.

ΔH= m(ΔH_{vap})

ΔH= 10g(2260) = 22,600 J

Finally: calculate the energy required to heat the now gaseous water those last 10 degrees.

ΔH = ΔT(C_{p})(m)

ΔH = [110-100](1.87)(10 g) = 187 J

Now, we take ALL of the ΔH's that we calculated and add them together:

206 J + 3340 J + 4,180 J + 22,600 J + 187 J = 30,513 J

But, the question asked for it in kJ, so dividing by 1,000 we get30.513 kJ. And that, my friend, is how much heat (in kJ) is required to warm 10 g of ice, initially at -10 deg. C, to steam at 110.0 deg C.

**pplantblues**- Posts : 109

Join date : 2008-11-14

## Re: Heat question

No problem. If in the future you have any more chem questions, feel free to post on this site.

**funion987**- Administrator
- Posts : 146

Join date : 2008-11-09

Similar topics

» Question on Lambda symbol for light

» Replacing heat sensor, problems with unplugging electronics

» Excessive heat loss from system

» Question re generational curses/ healing prayer

» Hot Water Heat Exchanger Pump on Constantly! Please Help!

» Replacing heat sensor, problems with unplugging electronics

» Excessive heat loss from system

» Question re generational curses/ healing prayer

» Hot Water Heat Exchanger Pump on Constantly! Please Help!

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum