Answer:
115 ⁰C
Explanation:
Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies
[tex]q_{1} +q_{2} =-q_{3}[/tex] -----eqution 1
where,
[tex]q_{1}[/tex] is the heat absorbed by the solid at 0⁰C
[tex]q_{2}[/tex] is the heat absorbed by the liquid at 0⁰C
[tex]q_{3}[/tex] the heat lost by the warmer water sample
Important equations to be used in solving this problem
[tex]q=m *c*\delta {T}[/tex], where -----equation 2
q is heat absorbed/lost
m is mass of the sample
c is specific heat of water, = 4.18 J/0⁰C
[tex]\delta {T}[/tex] is change in temperature
Again,
[tex]q=n*\delta {_f_u_s}[/tex] -------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
Step 2: calculate how many moles of water you have in the 100.0-g sample
[tex]=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O[/tex]
Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C
[tex]q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ[/tex]
This means that equation (1) becomes
79.13 KJ + [tex]q_{2} = -q_{3}[/tex]
Step 4: calculate the final temperature of the water
[tex]79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}[/tex]
Substitute in the values; we will have,
[tex]79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})[/tex]
79.13 kJ + 990.66J* [tex](T_{f}-218})[/tex] = -1463J*[tex](T_{f}-100})[/tex]
Convert the joules to kilo-joules to get
79.13 kJ + 0.99066KJ* [tex](T_{f}-218})[/tex] = -1.463KJ*[tex](T_{f}-100})[/tex]
[tex]79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3[/tex]
collect like terms,
2.45366[tex]T_{f}[/tex] = 283.133
∴[tex]T_{f} =[/tex] = 115.4 ⁰C
Approximately the final temperature of the mixture is 115 ⁰C
