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Heat question

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Heat question

Post  pplantblues on Fri Nov 14, 2008 10:46 pm

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?
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Re: Heat question

Post  funion987 on Sat Nov 15, 2008 1:42 am

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(Cp)(m)

ΔH= [-10-0](2.06)(10 g) = 206 J

Next: figure out the ΔHfus (enthalpy of fusion)
ΔH = m(ΔHfus)

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(Cp)(m)

ΔH = [100-0](4.18)(10 g) = 4,180 J

Next: calculate the ΔHvap for the liquid water turning into a gas.
ΔH= m(ΔHvap)

ΔH= 10g(2260) = 22,600 J

Finally: calculate the energy required to heat the now gaseous water those last 10 degrees.
ΔH = ΔT(Cp)(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.
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Thanks for your answer

Post  pplantblues on Sat Nov 15, 2008 2:33 pm

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(Cp)(m)

ΔH= [-10-0](2.06)(10 g) = 206 J

Next: figure out the ΔHfus (enthalpy of fusion)
ΔH = m(ΔHfus)

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(Cp)(m)

ΔH = [100-0](4.18)(10 g) = 4,180 J

Next: calculate the ΔHvap for the liquid water turning into a gas.
ΔH= m(ΔHvap)

ΔH= 10g(2260) = 22,600 J

Finally: calculate the energy required to heat the now gaseous water those last 10 degrees.
ΔH = ΔT(Cp)(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.
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Re: Heat question

Post  funion987 on Sat Nov 15, 2008 2:39 pm

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