Suppose I give you 2kg of a substance at 90 degree C with a specific heat of .75 cal/g degree C.

How much ice water will it take to cool the sample to 20 degree C?

To answer this question, we need to calculate the amount of heat energy the substance loses as it cools down, and then use that information to determine how much ice water is needed to absorb that heat.

To calculate the heat lost by the substance, we can use the equation:

Q = m * c * ΔT

Where:
Q is the heat lost (in calories)
m is the mass of the substance (in grams)
c is the specific heat of the substance (in cal/g degree C)
ΔT is the change in temperature (in degree C)

Given:
m = 2kg = 2000g (since 1kg = 1000g)
c = 0.75 cal/g degree C
ΔT = (90 - 20) = 70 degree C

Now let's substitute these values into the equation:

Q = 2000g * 0.75 cal/g degree C * 70 degree C
Q ≈ 105,000 calories

So, the substance loses approximately 105,000 calories of heat energy in order to cool from 90 degree C to 20 degree C.

Now, we need to determine how much ice water is required to absorb this amount of heat energy. For this, we need to know the specific heat of water and the heat of fusion for ice.

The specific heat of water is approximately 1 cal/g degree C, and the heat of fusion for ice is approximately 79.7 calories/g.

To find the amount of ice water needed, we can use the equation:

Q = m * c * ΔT + m * Hf

Where:
Q is the total heat energy (in calories)
m is the mass of the substance (in grams)
c is the specific heat of water (in cal/g degree C)
ΔT is the change in temperature (in degree C)
Hf is the heat of fusion of ice (in calories/g)

Since we are trying to find the mass of ice water, let's rearrange the equation and solve for m:

m = [Q - (m * c * ΔT)] / Hf

Substituting the known values:

m = [105,000 calories - (2000g * 1 cal/g degree C * 70 degree C)] / 79.7 calories/g
m ≈ 778 grams

Therefore, approximately 778 grams of ice water will be needed to cool the sample from 90 degree C to 20 degree C.