Answer:
115 ⁰C
Explanation:
<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies
-----eqution 1
where,
is the heat absorbed by the solid at 0⁰C
is the heat absorbed by the liquid at 0⁰C
the heat lost by the warmer water sample
Important equations to be used in solving this problem
, where -----equation 2
q is heat absorbed/lost
m is mass of the sample
c is specific heat of water, = 4.18 J/0⁰C
is change in temperature
Again,
-------equation 3
where,
q is heat absorbed
n is the number of moles of water
tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol
<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample
![=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O](https://tex.z-dn.net/?f=%3D237g%20%2A%5Cfrac%7B1%20mole%20H_%7B2%7D%20O%7D%7B18g%7D%20%3D%2013.167%20moles%20of%20H_%7B2%7DO)
<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C
![q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ](https://tex.z-dn.net/?f=q_%7B1%7D%20%3D%2013.167%20moles%20%2A6.01%5Cfrac%7BKJ%7D%7Bmole%7D%20%3D%2079.13KJ)
This means that equation (1) becomes
79.13 KJ + ![q_{2} = -q_{3}](https://tex.z-dn.net/?f=q_%7B2%7D%20%3D%20-q_%7B3%7D)
<u>Step 4:</u> calculate the final temperature of the water
![79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}](https://tex.z-dn.net/?f=79.13KJ%2BM_%7Bsample%7D%20%2AC%2A%5Cdelta%20%7BT_%7Bsample%7D%7D%20%3D-M_%7Bwater%7D%20%2AC%2A%5Cdelta%20%7BT_%7Bwater%7D)
Substitute in the values; we will have,
![79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})](https://tex.z-dn.net/?f=79.13KJ%20%2B%20237%2A4.18%5Cfrac%7BJ%7D%7Bg%5E%7Bo%7DC%7D%2A%28T_%7Bf%7D-218%7D%29%20%3D%20-350%2A4.18%5Cfrac%7BJ%7D%7Bg%5E%7Bo%7DC%7D%2A%28T_%7Bf%7D-100%7D%29)
79.13 kJ + 990.66J*
= -1463J*![(T_{f}-100})](https://tex.z-dn.net/?f=%28T_%7Bf%7D-100%7D%29)
Convert the joules to kilo-joules to get
79.13 kJ + 0.99066KJ*
= -1.463KJ*![(T_{f}-100})](https://tex.z-dn.net/?f=%28T_%7Bf%7D-100%7D%29)
![79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3](https://tex.z-dn.net/?f=79.13%20%2B%200.99066T_%7Bf%7D%20-215.96388%3D%20-1.463T_%7Bf%7D%2B146.3)
collect like terms,
2.45366
= 283.133
∴
= 115.4 ⁰C
Approximately the final temperature of the mixture is 115 ⁰C