The enthalpy of fusion of water is 6.01 kJ/mol. Sunlight striking Earth's surface supplies 168 W per square meter (1 W = 1 watt = 1 J/s). (b) The specific heat capacity of ice is 2.032 J/g°C. If the initial temperature of a 1.00 square emter patch of ice is -5.0°C, what is its final temperature after being in sunlight for 12 h, assuming no phase changes and assuming that sunlight penetration uniformly to a depth of 1.00 cm?

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Accepted Answer

Welcome back everyone to another video. The entropy of fusion of water is 6.01 kilojoules per mole. Sunlight striking earth surface supplies 168 of watts per square meter. And we also know that one of watt is equal to one joule per second. The specific heat capacity of ice is 2.032 joules per gram per Celsius. If the initial temperature of a one square meter patch of ice is negative five °C, what is its final temperature after being in sunlight for 12 hours? Assuming no face changes. And assuming that sunlight penetrates uniformly to a depth of a one centimeter. And we are given for answer choices. A 352 B 357 C 362 and D 347 all given in degrees Celsius. Now, this problem is not necessarily realistic, but it assumes no phase changes. So if we melt our eyes and we get liquid water, we're still assuming it's eyes. That's what it states. Meaning the only equation that we want to use here is Q equals cm multiplied by the change in temperature TF the final temperature minus the initial temperature T I. And now we can use this equation and essentially expand that right. Now, first of all, if we are solving for the final temperature, this is the goal of the problem, we can actually rearrange this equation and say, OK, if we want to solve for the final temperature TF, then we're taking heat dividing by cm and we're adding the initial temperature T I. So let's solve this problem. We have our setup right. We have our equation and that we simply want to solve it. So first of all, we want to determine the amount of heat. And for that purpose, we can, first of all understand that we have 168 watts, which we can express as joules per second. So we have 168 joules per second per square meter, right? So we're adding one second and 1 m squared on the bottom of the fraction. Now the area or basically the surface area of the block is one square meter. So we're multiplying by 1 m squared. OK. So this is what we're including in our calculation. OK. And now from here, we understand that if this is our products, we will end up with jewels per second, meaning to find the amount of heat in soda, we also need to multiply by time so that we eliminate seconds on the bottom. So let's go ahead and do that, we have 12 hours for our time. So we're multiplying by 12 hours and we want to convert that into seconds. So for that purpose, we're going to use the conversion pack which tells us that there are 3600 seconds in one hour. Ok. And now we understand that this gives us the amount of heat. Now, on the bottom of the fraction, we have the product cm. So first of all, we're using the specific heat capacity, 2.032 joules per gram per Celsius. And we need to multiply that by the mass of the block. Well, essentially to calculate the mass of the block, we need to use its volume and the density of ice, which is approximately in this case equal to the mass of I'm sorry, the density of water. So what we can do is take a 1 m squared, OK. That's what we have. And we understand that the depth is one centimeter. So if we are using centimeters, in this case, we can first of all convert 1 m squared into centimeters squared. So we're adding a meter squared on the bottom. And on top, we know that 1 m is equivalent to 10 to the power of second centimeters. So if we want to get centimeters squared, we're just squaring this number. Ok? And now what this tells us is that we are getting 10 of the four centimeters squared per 1 m squared. And if we multiply by 1 m squared, we noticed clearly that they canceled ch out and we are getting 10th of the fourth or 10,000 centimeters squared. And we're also multiplying by depth, right. So we're multiplying by one centimeter, this gives us volume and now we can multiply by the density because we're looking for mass. So we know that there's a 1 g per one centimeter cubed, that's the density of water. We're using it for the purposes of this problem. And now we are nearly done right. We have our heat on the top of the fraction, we have our product cm on the bottom of the fraction. And now we simply need to add the initial temperature which is negative 5.0 °C. And eventually we arrive at our final answer which is 352 degrees Celsius, which corresponds to the answer choice. A that would be our final answer. We notice that such a temperature already corresponds to the gas phase of water. But the problem states that we want to assume no phase changes. So that's why we didn't use the entropy of fusion. We didn't use the entropy of vaporization. We assumed that ice stays ice throughout this process. And that's why the problem is not necessarily really realistic. That would be our final answer. And thank you for watching.

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Key Concepts

Enthalpy of Fusion

Specific Heat Capacity

Energy Transfer from Sunlight