How much energy in Joules is released by burning a cheese ball that raises the temperature of

74.2 g water from 24°C to 41°C? How many Calories is this?

1. The heat energy in Joules is 5280.22J

2. The heat energy in calories is 1261.4 calories

Explanation:

Step 1:

Data obtained from the question.

The following data were obtained from the question:

Mass of water (M) = 74.2 g

Initial temperature (T1) = 24°C

Final temperature (T2) = 41°C

Change in temperature (ΔT) = T2 - T1 = 41 - 24 = 17°C

Heat energy (Q) =?

Recall:

Specific heat capacity of water = 4.186 J/g°C

Step 2:

Determination of the energy in Joules.

This is illustrated below:

Q = MCΔT

Q = 74.2 x 4.186 x 17

Q = 5280.22J

Step 3:

Conversion of the heat energy in Joules to calories. This is illustrated below:

4.186 J = 1 calorie

Therefore, 5280.22J = 5280.22/4.186 = 1261.4 calories

dynamic equilibrium is the answer.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :