Question

Given 319.5 g of hot tea at 72.0 ∘C, what mass of ice at 0 ∘C must be added to obtain iced tea at 13.5 ∘C? The specific heat of the tea is 4.18 J/(g⋅∘C), and ΔHfusion for ice is +6.01 kJ/mol.

Answer 1

The formula for these types of problems is M1 x C1 x dT1 = M2 x C2 x dT1
M₁ is mass for iced tea =

M₂ is massfor the ice cube
C₁ is specific heat capacity of tea= 4.18J/g ^oC

C₂ is specific heat of ice = 2.09 J/g°C
dT is temperature difference for the iced tea it's 59.5 degrees, and for the icecube its 13.5 degrees.

This is a tricky problem though, because the ice cube will actually melt in the process.
You can't use the formula like this then.

The best thing to do is assume the ice is actually water at a much lower temperature.
imagine that water never freezes.

First, we heat the ice 13.5 degrees.
Ice's specific heat capacity is only 2.09 J/g^oC though, while water's is 4.18 J/g^o​C
Therefore,
Just imagine you're heating water 5.643 degrees.

Next, find the energy required to melt the ice.
They gave the latent heat as 6.01 kJ/mol, but we need it in grams.
Since water weighs 18 amu, just divide it by 18
6.01 / 18 = .33888...
A kilojoule is 1000 joules, so multiply it by 1000\
338.888... joules to melt 1 gram of ice.
Since water's specific heat capacity is only 4.18 joules to heat it 1 degree,
divide 338.888 / 4.18 = 79.8 degrees

Now add up those 79.8 degrees for melting the ice, and 6.75 degrees for heating it.
86.25 degrees below 0.

Imagine the water is -86.25 degrees.

319.5×4.18×59.5 = M₂×4.18×13.5

M₂ = 1408.16gm

The answer is you need 1408.16gm grams of ice to cool down the water to 10 degrees