In lab, a student burns 0.550 grams of Ritz crackers. The burning causes 100.0 grams of water to rise from 24.0°C go 47.1 °C.

Calculate the number of calories absorbed by the WATER as the Ritz crackers burn. (Use the specific heat of water in terms of calories: 1.0 cal/ g °C) ** Remember you are calculating the heat gained by the water, NOT the heat lost by the Ritz cracker - use the correct mass!!

12.7 cal

53.2 cal

2310 cal

9.67 x 105 cal

2310 calories are absorbed by water to raise its temperature.

Option 3.

Explanation:

For this problem, we can use the specific heat formula for determining the heat energy required for this process. The heat energy required for this process will be equal to the number of calories absorbed by the water.

So, the formula is

Q = mcΔT

Here, we have to use the mass of water which has to rise its temperature, so the m = 100 g. Similarly, the specific heat of water should be taken into consideration, so c = 1 cal/g°C.

Then the change in temperature ΔT = 47.1-24 = 23.1 °C

So, the calories absorbed by the water will be

Thus , 2310 calories are absorbed by water to raise its temperature.

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93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.