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Ch.1 - Chemical Tools: Experimentation & Measurement

Chapter 1, Problem 79

Assume that the kinetic energy of a 1400 kg car moving at 115 km/h (Problem 1.78) is converted entirely into heat. How many calories of heat are released, and what amount of water in liters could be heated from 20.0 °C to 50.0 °C by the car's energy? (One calorie raises the temperature of 1 mL of water by 1 °C)

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Hello everyone today. We have the following problem. A bus traveling north has a kinetic energy of 2.8 times 10 to the sixth jewels. If half of the kinetic energy is converted into heat, how much heat in calories are released and how many grams of water could be heated by this converted energy from 25 degrees Celsius to 45 degrees Celsius. And then we have the specific heat capacity of water which is 4. jewels per grams Celsius. So first we need to recall that one calorie is equal to 4.184 jules. And now we need to calculate the heat that's released so the heat releases in unit. It's going to be represented by Q. It was half of the kinetic energy And we're going to multiply that by the number of jewels and then by our conversion factor. So the kinetic energy in jewels is 2.8 times 10 to the sixth jewels. And that we need to get rid of the jewels to find the calories. So we use the conversion factor, that one calorie Is equal to 4.184 jules. Our units of jewels cancel out and we are left with 3.3 times 10 to the fifth calories. So that's how many calories are heat. That's how much heat in calories is released. And so we need to find the grams of water. First we're going to use our specific heat formula which is Q. Equals the mass times the specific heat times our change in temperature. When we rearrange this, we get that the mass is going to be equal to our heat over the specific heat times the change in temperature. So we already calculated The value for Q, which was 3.3 times 10 to the 5th cows. Our specific heat capacity was said to be 4. jules, programs, Celsius. And then our change in temperature. We went from Our final temperature was 45°C and our initial temperature was 25°C. And when we solve this, we get 1.7 times 10 to the 4th g of water as our final answer. And so these are going to represent our two answers for our question. Overall, I hope that this helped and until next time.