Question

Part A) A 1401 kg car is traveling down the road at 83.4 km/h. While traveling at this rate of speed, what is the kinetic energy of this vehicle in kilojoules?

Part B)The specific heat of a certain type of cooking oil is 1.75 J/(g·°C). How much heat energy is needed to raise the temperature of 2.79 kg of this oil from 23 °C to 191 °C?

Part C) A researcher studying the nutritional value of a new candy places a 5.40-gram sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.21 °C. If the heat capacity of the calorimeter is 33.90 kJ·K–1, how many nutritional Calories are there per gram of the candy?

Answer 1

a)

Given :-

v = 83.4 km/h = 23.16667 m/s

m = 1401 kg

E = 1/2*mv^2

E = 0.5 x 1401 kg x 23.16667^2

E = 375954.56 J

E = 375.95 KJ

B)

mass of a cooking oil, m = 2.79 kg = 2.79 x 10^3 g

specific heat of a cooking oil, 1.75 J/(g·°C)

change in temp, 191 - 23 = 168 °C

Heat energy, q = m*s*t

q = 2.79 x 10^3 g x 1.75 J/(g·°C) x 168 °C

q = 820260 J

q = 820.260 KJ

C)

quantity of heat = heat capacity x change in tempreture

= 33.90 kJ·K–1 x 2.21 °C

= 74.919 KJ

heat = 74.919 KJ / 5.40 g = 13.87 KJ/g

heat = 13.87 x 1000 J / 1g x 1 cal / 4.2 J x 1 cal / 1000 cal

heat = 3.30 cal