If there are 160 Calories in 28 g in 1 serving of Flamin' Hot Cheetos, how many gram(s) of Cheetos must you burn to raise the temperature of 20.0 mL of water 1oC?

3.5 g

Explanation:

The density of water is 1 g/mL. The mass (m) corresponding to 20.0 mL is 20.0 g.

We can calculate the heat (Q) required to raise the temperature of 20.0 mL of water 1 °C (ΔT).

Q = c × m × ΔT = 1 cal/g.°C × 20.0 g × 1 °C = 20 cal

where,

c is the specific heat capacity of water

There are 160 calories in 28 g of Cheetos. The mass that releases 20 cal is:

20 cal × (28 g/160 cal) = 3.5 g

im doing test rn im going with 2 but ill give answer if itss wrong,   : )