When a 0.66-g sample of olive oil is burned in a calorimeter, the...

Question

When a 0.66-g sample of olive oil is burned in a calorimeter, the heat released increases the temperature of 370 g of water from 22.7 °C to 38.8 °C. What is the energy value for the olive oil in kcal/g? (3.5, 3.6)

Accepted Answer

Welcome back, everyone. Calculate the energy value in cows per gram. If the temperature of 380 g of water increased from 25 Celsius to 40.1 Celsius. 1.80 g of butter was burned in a calorimeter. We're going to need to recall our formula for specific heat capacity to figure out the heat capacity of the water in our calorimeter, which will ultimately lead us to getting the energy value of our butter when it's burned in the calor meter. So we call that our formula describes our specific key capacity represented by the variable C, which is lower case, this is set equal to our value for heat represented by the variable Q and divided by our mass M multiplied by the difference in temperature. Delta T. And this is with regard to our substance. In this case, we're then going to want to find the heat content of our water in the calor meter in order to find the energy value of our butter. So we need to isolate for Q and we're going to do so by multiplying both sides by our denominator M times delta T, so that it will cancel out on the right. And we would be left with the fact that our heat Q is found from taking our specific key capacity times our mass of our substance times the difference in temperature of our substance. So finding the heat of our water in our calor meter, we would set this equal to our specific key capacity of water, which from our textbooks, we should recall is the value 1.0 calories per gram times degrees Celsius. Then multiplying by our massive water given in our prompt as 380 g and then multiplying by delta T where we have 20 or sorry, our final temperature given as 40.1 degrees Celsius subtracted from our initial temperature of 25 degrees Celsius. In our next line, we wanna cancel our units of degrees Celsius as well as grams leaving us with calories and we should have 1.0 calories multiplied by 380 multiplied by the result of our difference between our temperatures being 15.1. So in our next line, taking the product between these three values, we'll get a value of 5738 calories. However, because we need to give our energy value of our butter and kill a calories per gram. We're going to convert our heat of water to killer calories by multiplying by the conversion factor to go from calories in the denominator to kill a calories in the numerator where we recall that our prefix kilo tells us that we have an equivalent of 10 to the third of our base unit K, canceling out calories were left with kilocalories. And we'll find that our heat of our water was equal to 5.738 kg. Now that we have the heat content of our water, we're going to find our energy value and kill a cow's per gram of the butter in the calor. So where we would ultimately find this by taking the heat content, Q divided by our mass of our butter. So this would equal again, our value for Q, we found to be 5.738 kg. And we're dividing by the mass of our butter given in the prompt as 0.80 g of butter. So this will simplify. So that from our quotient, we have a result of 7.1725 kg per gram as our energy value of our butter. And because our calculation from our quotient had the smallest number of decimal places being two decimal places given in our mass of butter. We want around our final answer to two sig figs as 7.2 kill cows per gram as our energy value of our butter. So 7.2 K cos per gram is our final answer. I hope this made sense and let us know if you have any questions

Struggling with GOB Chemistry?

When a 0.66-g sample of olive oil is burned in a calorimeter, the heat released increases the temperature of 370 g of water from 22.7 °C to 38.8 °C. What is the energy value for the olive oil in kcal/g? (3.5, 3.6)

Watch next

Nature of Energy practice set