Question

Find the resulting temperature when 100 g of ice at 0oC is mixed with 1 liter of H2O at 35oC. 

Answer 1

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

where

q₁ - the heat absorbed by the solid at 0°C

q₂ - 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 × ∆T, where

q - heat absorbed/lost

m - the mass of the sample

c - the specific heat of water, equal to 4.18 J/g°C

ΔT - the change in temperature, defined as final temperature minus initial temperaturea and

q = n × ∆H_fus, where

q - heat absorbed

n - the number of moles of water

ΔH_fus - 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

100.0 g × 1 mol H₂O / 18.015 g = 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?

q = 5.551 moles × 6.01 kJ/mole = 33.36 kJ

This means that first equation becomes

33.36 kJ + q₂ = −q₃

The minus sign for q₃ is used because heat lost carries a negative sign.

So, if T_f is the final temperature of the water, you can say that

33.36 kJ + m_sample × c × ΔT_sample = −m_water × c × Δ

T

water

More specifically, you have

33.36 kJ + 100.0 g × 4.18J/g°C × (T_f − 0)°C = −1000 g × 4.18 J/g°C × (T_f − 35)°C

33.36 kJ + 418 J × (T_f − 0) = −4180 J × (T_f − 35)

Convert the joules to kilojoules to get

33.36 kJ + 0.418 kJ × T_f = −4.180 kJ × (T_f − 35)

This is equivalent to

0.418 kJ × T_f + 4.180 kJ × T_f = 146.3 kJ − 33.36 kJ

T_f = 24.56°C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T_f = 25°C