# How much ice (in grams) would have to melt to lower the temperature of 355mL of water from 26∘C to 6∘C? (Assume the density of water is 1.0 g/mL.)

## My solution assumes the the ice is at zero C.

## Well, it seems like you're trying to make some icy calculations! Let's get down to it. The change in temperature is 20∘C (26-6). To lower the temperature of water by 1∘C, we need to melt 1 gram of ice, as per the assumption that you provided. So, to lower the temperature by 20∘C, we would need to melt...you guessed it...20 grams of ice!

## To answer this question, we need to understand the concept of specific heat and the heat equation.

The heat equation is:

Q = m * c * ΔT

Where:

Q is the heat energy transferred

m is the mass

c is the specific heat

ΔT is the change in temperature

In this case, we want to find the amount of ice (in grams) that would have to melt to lower the temperature of water from 26∘C to 6∘C.

First, let's calculate the amount of heat energy needed to lower the temperature of the water.

To do this, we need to calculate the heat energy required to lower the temperature from 26∘C to 0∘C and the heat energy to melt the ice.

1. Heat energy to lower the temperature from 26∘C to 0∘C:

Q1 = m1 * c * ΔT1

Given:

m1 = 355 mL (water)

c = 1.0 g/mL (specific heat of water)

ΔT1 = 26∘C - 0∘C = 26∘C

Q1 = m1 * c * ΔT1

Q1 = 355 g * 1.0 g/mL * 26∘C

2. Heat energy to melt the ice:

Q2 = m2 * ΔHf

Given:

ΔHf = heat of fusion of ice = 334 J/g

We need to convert the heat energy calculated in step 1 to joules because the heat of fusion is in joules per gram.

Q1 (in joules) = Q1 (in grams) * specific heat of water (4.184 J/g⋅∘C)

Finally, we can calculate the mass of ice that would have to melt to lower the temperature from 26∘C to 6∘C.

Q = Q1 + Q2

Given Q (the total heat energy) and ΔHf (the heat of fusion), we can solve for m2 (the mass of ice):

Q = m2 * ΔHf

m2 = Q / ΔHf

Substituting the values of Q1 into this equation will give us the mass of ice required to lower the temperature of 355mL of water from 26∘C to 6∘C.

## 355*4.18*(6-26)=-29678

## heat lost by ice + heat gained by melted ice + heat lost by water @ 20 cooling to 6 = 0

(x g ice x heat fusion) + [(x g melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass 20C H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

Solve for x grams ice.

The answer I obtained is approx 80 g.