An iron ball of mass 100g is heated to 100⁰C and is put on a block of ice at 0⁰C. If 12.5g of ice melts, specific heat of iron is...( take L = 80 cal/g)
Answers
Answer:-11∘C
Explanation:
As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at
0°C to liquid at 0∘C
The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have
q1+q2=−q3 (1), where
q1 - the heat absorbed by the solid at 0∘C
q2- the heat absorbed by the liquid at 0∘C
- the heat absorbed by the liquid at 0∘C
q3 - the heat lost by the warmer water sample
The two equations that you will use are
q=m⋅c⋅ΔT , where
q - heat absorbed/lost
m - the mass of the sample
c - the specific heat of water, equal to
4.18J/g∘C
ΔT - the change in temperature, defined as final temperature minus initial temperature
and, q=n⋅ΔHfus,
where,
q - heat absorbedn- the number of moles of water
ΔHfus - the molar heat of fusion of water, equal to
6.01 kJ/mol
Use water's molar mass to find how many moles of water you have in the 100.0-g sample
100.0g⋅1 mole H is 2O18.015g=5.551 moles H2O
So, how much heat is needed to allow the sample to go from solid at 0∘C to liquid at 0∘C ?
q1=5.551moles⋅6.01kJmole=33.36 kJ
This means that equation
(1) becomes 33.36 kJ+q2= −q3
The minus sign for
q3 is used because heat lost carries a negative sign.
So, if Tf is the final temperature of the water, you can say that
33.36 kJ+msample⋅c⋅ΔTsample = −mwater⋅c.ΔTwate
More specifically, you have
33.36 kJ+100.0g⋅4.18Jg∘C⋅(Tf−0)∘C= −650g⋅4.18Jg∘C⋅(Tf−25)∘C33.36 kJ+418 J⋅(Tf−0)=−2717 kJ⋅(Tf−25)
Convert the joules to kilojoules to get
33.36kJ+0.418kJ⋅Tf= −2.717kJ⋅(Tf−25)
This is equivalent to
0.418⋅Tf+2.717⋅Tf=67.925−33.36
Tf=34.5650.418+2.717= 11.026∘C
Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be
Tf = 11°C
Answer.....
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