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.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.