When a sample of food is burned, it caused the temperature of the water in a calorimeter containing 1990. g of water to increase by 8.9oC.

Question

I also want to know. What is the mass of the sample if it was 100% carbohydrate? (carbs =4.0 kcal/g)

Answer 1

Hello!

Let's start this question off by remembering the equation for calorimetry and the specific heat of water:

q = mcΔT

where q is the heat in energy terms, m is the mass in grams, c is the specific heat which can be expressed in a few different ways, and ΔT is the change in temperature in °C or K (I'll be just be using °C going forward, but it works for both units).

Energy can be defined in Joules (J) or calories (cal), but given the information given to us about how to get the mass of carbohydrate from energy burned, let's use kcal.

The specific heat of a substance describes how much energy it takes to raise the temperature of one gram of material by 1 °C. The specific heat of water is 0.001 kcal per gram per °C.

Let's now identify our variables and plug them into the equation:

m = 1990. g

c = 0.001 kcal per gram per °C

ΔT = 8.9 °C

q = mcΔT

q = (1990.)(0.001)(8.9)

q = 17.711 kcal

Now that we have our energy in kcal and a rate to go from kcal to grams of carbs, we can get our final answers using unit conversion:

17.711 kcal*[(1 g)/(4.0 kcal)] = 4.42775 g carbs

Finally, keeping our significant digits in mind we have our final answer of 4.4 g carbs.

I hope this clarifies calorimetry problems a bit more for you!