Question

A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect world all of the energy from burning the food will be transferred from the food burning to the water in the calorimeter. Assuming no energy loss, what would be the final temperature of the water if you began with 23 L of water at 16 oC?

Answer 1

38.3958 °C

Step-by-step explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories of energy

1 gram of protein on burning gives 4 kilocalories of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy

Total energy = 515 kilocalories

Using,

`Q=m_(water)* C_(water)* (T_f-T_i)`

Given: Volume of water = 23 L = 23×10⁻³ m³

`Density=(Mass)/(Volume)`

Density of water= 1000 kg/m³

So, mass of the water:

`Mass\ of\ water=Density * {Volume\ of\ water}`

`Mass\ of\ water=1000 kg/m^3 * {0.023\ m^3}`

Mass of water = 23 kg

Initial temperature = 16°C

Specific heat of water = 0.9998 kcal/kg°C

`515=23* 0.9998* (T_f-16)`

Solving for final temperature as:

Final temperature = 38.3958 °C