An ice cube with a mass of 20 g at -20 °C (typical freezer temperature) is dropped into a cup that holds 500 mL of hot water, initially at 83 °C. What is the final temperature in the cup? The density of liquid water is 1.00 gmL; the specific heat capacity of ice is 2.03 Jg@C; the specific heat capacity of liquid water is 4.184 Jg@C; the enthalpy of fusion of water is 6.01 kJmol.

Question

An ice cube with a mass of 20 g at -20 °C (typical freezer temperature) is dropped into a cup that holds 500 mL of hot water, initially at 83 °C. What is the final temperature in the cup? The density of liquid water is 1.00 g>mL; the specific heat capacity of ice is 2.03 J>g@C; the specific heat capacity of liquid water is 4.184 J>g@C; the enthalpy of fusion of water is 6.01 kJ>mol.

Accepted Answer

Hey everyone in this example we have 17 g of a negative 4.5 degree Celsius ice cube. Put into a beaker with 350 mL of 78 degrees Celsius. Hot water. We need to calculate the final temperature in the beaker and we're told that the entropy of fusion of water is this value here as well as its specific heat capacity equal to this value here and were given the specific heat capacity of ice here. Lastly we're also given the density of water as 1.0 g per middle leader. So because our ice cube is placed in our beaker of hot water, we can say that our heat lost by the hot water in the cup is equal to the heat gained by our ice cube. So for our 17 g ice cube we are going to multiply this by the heat capacity of ice given in the prompt as 2.3 jewels divided by grams times degrees Celsius. And then we're going to also multiply this by our difference between the temperature which is given the prompt as negative 4.5 degrees Celsius. So what we would get here is a value equal to 155,000 0.295 jewels. And this is a negative value. However, because this is corresponding to heat gained, we're going to just interpret this as a positive value because energy can only be positive and now we want to move on to our second part of our solution where we take our 17 g of our ice cube and we're going to multiply it by the molar mass of water where we recall that for one mole of water we have 18.02 g. And sorry about that, we have 18.02 g of water. We are then going to multiply by the entropy of fusion of water which is given in the prompt as 6.1 kg joules per mole of water. Now we want our units here to end up in jewels for energy. So we're going to convert from kilo jewels, two jewels. And recall that our prefix kilo tells us we have 10 to the third power jewels. So this is going to give us a value equal to 56,000. We're sorry 5669.8113. And we have units of jewels because all of our units are going to cancel out kayla jewels and molds will cancel out as well as Graham's. Likewise, with our above calculation, we had Celsius cancel out and Graham's leaving us with jewels for both of our energy values. Next we want to take our mass of our ice and we want to multiply it by our specific heat capacity of water given in the prompt as 4.184 jewels divided by grams times degrees Celsius. And we're going to multiply it by our final temperature of our beaker, which we need to solve for subtracted from the initial temperature of our ice. So this is going to give us a result of 71. where we have the final temperature still subtracted from 0°C. Now last we're going to take our volume of our beaker, 350 mL of our hot water. And we're going to convert first from milliliters two g by recalling for one millimeter we have one g And then we're going to multiply this by the specific heat capacity of water which given in the prompt is 4. jewels divided by g times degrees C. And then we're going to multiply this by our final temperature. Or rather by the initial temperature of our beaker given in the prompt as 78°C subtracted from the final temperature of our beaker with the ice. And this is going to give us a result equal to 1,464.4. Multiplied by 78°C subtracted from the final temperature. And as far as our units, we were able to cancel out degrees Celsius, we were able to cancel out grams and milliliters. Likewise above, we were able to cancel out grams degrees Celsius and we were left with jewels in both cases. So we can also write in J for jewels here. So to get to this fourth equation here, we had to take parts one through two through three and add them together. So we would say that this is 12 and three and this led us to part four. So we're going to say that equation four, which is Which is a result of 1,464.4 jewels times 78°C is going to equal our result for parts one through three. So for part one we have a value of 155.2 95 jewels. For part two we're going to add this as point or sorry, part two was 56, or 5669. jewels. And then for part three we got 71.1 multiplied by the final temperature minus zero degrees Celsius. So we want to simplify this to solve for the final temperature. To complete this example. So what we would do is distribute on the left hand side so that we can simplify to 114223.2 subtracted from 1464.4 times the final temperature. This is set equal to our right hand side where we take the sum of the first two values. So we have and sorry, this is a five here .1063. This is in units of jewels. This is then added to 71.128 times the final temperature. So simplifying for just the final temperature, we can say that the final temperature is equal to a value of 70. degrees Celsius. So this will complete this example as our final answer. I hope that everything I reviewed was clear, but if you have any questions, just leave them down below and I will see everyone in the next practice video.

Verified Solution

Key Concepts

Specific Heat Capacity

Enthalpy of Fusion

Heat Transfer and Equilibrium