CALC You make tea with 0.250 kg of 85.0°C water and let it cool to room temperature 120.0°C2. (a) Calculate the entropy change of the water while it cools.

Video transcript

Welcome back everybody. We are taking a look at a golden ring that is in the care of a jeweler. We are told that the mass of this gold ring is eight g and that the jeweler heats it to 500 degrees Celsius. Then he lets it naturally cool all the way down to 30 degrees Celsius. Now, since we're working with golden heat, it would be good to know. The specific heat of gold is 0.128. Uh sorry jules, jules per gram kelvin. And we are tasked with finding what the change in entropy is. As the golden ring cools well, the formula for the change in entropy is going to be the mass times the specific heat times the natural log of the final temperature divided by the initial temperature. But before we can start plugging in values, we need to make sure everything is In the correct units. And specifically our temperatures need to be in Kelvin. So I'm going to convert them to Kelvin real quick by adding 273.15 to each giving us 81 of 773.15 Kelvin and 82 of 303.15 Kelvin. Great. Now we are ready to plug in our numbers here we have that delta S. Is equal to eight g times 0.128 jewels per gram Calvin times the new Natural log of our final temperature 303.15 Kelvin divided by our initial temperature of 773.15 Kelvin at the beginning of the cooling process, canceling out some units. Here, we are left with our desired units of joules per kelvin, and specifically when we plug this into our calculator, we get a change in entropy of negative 0.96 joules per kelvin, which corresponds to our answer choice of B. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.