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.

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

-----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

Step 2: calculate how many moles of water you have in the 100.0-g sample

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

This means that equation (1) becomes

79.13 KJ +

Step 4: calculate the final temperature of the water

Substitute in the values; we will have,

79.13 kJ + 990.66J* = -1463J*

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* = -1.463KJ*

collect like terms,

2.45366 = 283.133

∴ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

3.2x10^7