Question

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answer 1

115 ⁰C

Step-by-step explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

`q_(1) +q_(2) =-q_(3)` -----eqution 1

where,

`q_(1)` is the heat absorbed by the solid at 0⁰C

`q_(2)` is the heat absorbed by the liquid at 0⁰C

`q_(3)` the heat lost by the warmer water sample

Important equations to be used in solving this problem

`q=m *c*\delta {T}`, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

`\delta {T}` is change in temperature

Again,

`q=n*\delta {_f_u_s}` -------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

`=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O`

Step 3: 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(KJ)/(mole) = 79.13KJ`

This means that equation (1) becomes

79.13 KJ +
`q_(2) = -q_(3)`

Step 4: calculate the final temperature of the water

`79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)`

Substitute in the values; we will have,

`79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})`

79.13 kJ + 990.66J*
`(T_(f)-218})` = -1463J*
`(T_(f)-100})`

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ*
`(T_(f)-218})` = -1.463KJ*
`(T_(f)-100})`

`79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3`

collect like terms,

2.45366
`T_(f)` = 283.133

∴
`T_(f) =` = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C