Using the values for the heat of fusion, specific heat of water, ...

Question

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:c. kilojoules needed to melt 24.0 g of ice at 0 °C, warm the liquid to 100 °C, and change it to steam at 100 °C

Accepted Answer

Welcome back, everyone. Using the following values, our entropy of fusion of ammonia equal to 332.3 Jews per grams. Our specific heat capacity of ammonia in liquid form equal to 4.744 jules per grams times degrees Celsius. And our entropy of vaporization of ammonia equal to 1370 Jews per gram, determine the amount of heat energy and kilojoules needed to convert a 50 g sample of solid ammonia at negative 77.73 degrees Celsius to liquid ammonia. And we're going to heat the liquid to negative 33.33 degrees Celsius and change the liquid gas, the liquid to gas. Sorry at 33 negative 33.33 degrees Celsius. Now we're going to set three variables of heat to represent our conversions of ammonia as far as its phase conversions. So we can set a plot where we would compare temperature on our Y axis to time on our X axis and utilizing our temperature range. Given in our prompt, we would begin at the origin with zero degrees Celsius. We would have a mid-range temperature of around negative 33.33 degrees Celsius as given in the prompt. And then we have our lowest temperature of negative 77.73 degrees Celsius given from the prompt. Now we're going to set three variables to represent our heat conversions of ammonia. And we will utilize the variables Q one, Q two and Q three for Q one, let's utilize this to represent our conversion of ammonia from its solid phase to its liquid phase. And this would be related to our entropy of fusion of ammonia, which is given to us in our prompt. From the prompt. We're told that to convert 50 g of solid ammonia at negative 77.73 degrees to liquid ammonia. We're going to actually utilize that information and plug that Q one variable around our temperature of negative 77.73 degrees Celsius. So we have Q one at this point. Now let's focus on our second variable Q two where we can utilize this variable to represent our heat energy that is required to raise our temperature from negative 77.73 degrees Celsius to our higher value of negative 33.33 degrees Celsius. And so therefore, we would want to utilize our specific heat capacity represented by the term C and for ammonia that is given to us for liquid ammonia being 4. jules programs times degrees Celsius. So we can note that down later. Now we can plug in Q two at our higher value of temperature negative 33.33 degrees Celsius on our plot. So around this point, so connecting Q one to Q two so far, we would have the following curve. And now we are going to focus on Q three, our third variable in which describes the heat energy to change our liquid form of ammonia to its gaseous form. And this will occur at negative 33.33 degrees Celsius. And this is where we can utilize our third given term, which is our entropy of vaporization of ammonia, which we can use for phase changes, which will occur when we go from liquid to gas. So, plugging in Q three, we actually want our gas form of ammonia to be at negative 33.3 degrees Celsius. So we're actually going to lower Q two so that it's at a mid point on our curve. And then we would be able to place Q three at negative 33. degrees Celsius since that is when ammonia is within its gas form. So labeling our curb, we would have solid ammonia to liquid ammonia at our points between Q one. Then for Q two, we would have our heating of liquid ammonia up to negative 33. degrees Celsius where it's finally transitioned from liquid ammonia to its gaseous form. So now we need to determine our actual values for Q one through Q three. Beginning with Q one, let's start with Q one where we'll assign it the color red. We're going to begin by taking our mass of ammonia and we want to multiply it by our entropy of fusion of ammonia. And so this would describe our face change of ammonia from its solid to liquid form in which we plug in our mask given as 0.0 g of ammonia. And we're going to multiply bar alibi of fusion of ammonia to change from solid to liquid form, which is given to us in the prompt as 332.3 Jews per gram. Again, the heat required to make that phase change. Now canceling our units of grams. Notice that we're left with our unit of jewels, which is what we would need for A Q in terms of heat. And this will simplify so that our value for Q one is equal to 1.6615 times 10 to the fourth power. And we're left with units of Jews. So this is our value for Q one. Let's keep going. So we can get Q two and Q three so that we can calculate our total heat energy and kilojoules to make all of our phase changes between Q one through Q three. So moving on to Q two, we'll use the color green. Now. So for Q two, we're going to take our mass of ammonia multiplied by our specific heat capacity of ammonia, which is then multiplied by our difference in temperature when it's converted from its or when it's heated to negative 33. degrees Celsius before converting to a gas form. So, plugging in our mass, we have 50.0 g of ammonia multiplied by its specific he capacity given in the prompt. Our term C as 4.744 Jews per gram times degrees Celsius. So just making more room for our delta T term recall that that takes our final temperature which would be the higher temperature of negative 33.33 degrees Celsius minus our initial temperature of ammonia in its liquid form being negative 77.73 degrees Celsius. So since we have a negative and a negative, we're gonna add these two terms in our temperature. And in our next slide, we will simplify so that we have 50.0 g of ammonia. And actually, we can just take the product in the beginning so that we have a product of 237.2. Notice that our units of grams will cancel as well as our units of Celsius leaving us with jewels. So we have 237.2 Jews which is multiplied by our results from our Delta T being 44.4, taking the product between our two values will get our final result as our total heat for heating our liquid ammonia being 10531.7. Jos again, this is our total heat to heat liquid ammonia to a temperature of negative 33. degrees Celsius. And now we're going to find our term for Q three, which is describing our phase change of liquid to gaseous ammonia. So to find our variable Q three, we would take our mass of ammonia and multiply by our entropy a vaporization of ammonia to change from liquid form to gaseous form. And this value is given to us. So plugging in our mask first, we have 50.0 g of ammonia, which is multiplied by its entropy of vaporization given in the prompt as 1370 jewels per gram. So canceling out our unit of grams, we're left with again units of jewels for our heat. And we would find that our total heat Q three is equal to 68,500 Jews. Now to find our total heat, which we say is Q total to undergo all of our conversions from our phase changes between Q one through Q three, we would take our sum between Q one to Q two and Q three. So plugging in our values from above for Q one, we found a value of 1.66 15 times 10 to the fourth power jewels. Now because this was written in scientific notation, let's take our value for Q two and also write it in scientific notation by moving our decimal point over to the left by four places so that we have a value of 1.532 times 10 to the positive fourth power jewels. So plugging that in, we have plus 1.532 times 10 to the fourth power Jess and also writing our value for Q three in scientific notation. We would also move our decimal 30.4 places so that we have 6.85 times 10 to the positive fourth power Jews. So adding that in, we're then going to plug in our sum in our calculator, which will yield a value for Q total of 9500 or 95,647 Jews. However, writing this in scientific notation and rounding up, we're going to move our decimal point over four places so that we have this equal to about nine point 565 times 10 to the fourth power duels as our total heat required to make our conversions between Q one to Q three. Now we're going to need to convert this to kilojoules. So we're going to multiply it by our conversion factor to go from jewels in the denominator to kilojoules in the numerator where we recall that our prefix kilo tells us we have an equivalent of 10 to the negative third of our base unit jewel in the denominator. So canceling out Jews, we're left with Killer Jes as our final unit for our total heat and we would find our total heat equal to and apologies just to correct our conversion. This should be a positive power of three. So one kilojoule equivalent to 10 to the third Jews in the denominator. And then we would have our final accurate answer of 95.65 kilojoules of heat in total as our final answer of heat required and kill the Jews to convert our 50 g sample of solid ammonia from its liquid form and then from its liquid form to its gaseous form. So I hope that this made sense and let us know if you have any questions.

Struggling with GOB Chemistry?

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:c. kilojoules needed to melt 24.0 g of ice at 0 °C, warm the liquid to 100 °C, and change it to steam at 100 °C

Watch next

Heating and Cooling Curves practice set