30 g of copper pellets are removed from a 300°C oven and immediately dropped into 100 mL of water at 20°C in an insulated cup. What will the new water temperature be?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with heat transfer. Let's see what it's asking us. The scientist extracts 220 g of aluminum from a furnace at 400 degrees Celsius and places them into 150 mL of water that's held at 15 degrees Celsius. We're asked to determine the resulting equilibrium temperature after the um aluminum nuggets are. So our multiple choice answers are going to be in units of degrees Celsius. A 107 B 92.7, C 101 or D 97.2. So the first thing we can do here is recognize that because we are working in a thermally insulated container, we have a conservation of energy. So we have the heat energy from the aluminum plus, the heat energy from the water is going to equal zero. We can recall that our heat energy term is given as MC delta T. So we'll have the mass of the aluminum multiplied by C the specific heat capacity of aluminum multiplied by our delta T of the aluminum, which I'm going to write as T F minus T initial. In this case, T alumna initial, our T F is going to be the same for both the aluminum and the water because it is our equilibrium temperature. So similarly, for our water heat energy, we have plus the mass of the water multiplied by the specific heat capacity of the water multiplied by our final temperature minus our initial temperature of the water. And so when we isolate out that final temperature, turn, we have our final temperature multiplied by the mass of the aluminum multiplied by the specific heat capacity of aluminum plus the mass of the water multiplied by the specific heat capacity of the water. On the left hand side of the equation, and then that equals the mass of the aluminum multiplied by the specific heat capacity of the aluminum multiplied by the initial aluminum temperature plus the mass of the water, specific heat capacity of the water and initial water temperature. And so when we look at each of these terms and we're trying to solve for T F. So the mass of the aluminum was given in the problem as 220 g to keep us in standard units. I'm going to rewrite that as 2200.22 kg multiplied by our specific heat capacity of the aluminum, which we can recall is 900 Jews per kilogram Celsius. And then the initial temperature of the aluminum was given in the problem as 400 degrees Celsius. We're gonna add that to our initial water term. So the mass of the water, we can recall that our density of water is related to the mass of water by one kg is equal to one liter. And so when we have 10.15 liters, that is equal to 0.15 kg. And then multiply that by the specific heat capacity of the water, we can recall that's 4186 Jews per kilogram. C multiplied by that initial temperature of water which is 15 degrees Celsius. All of that divided by the mass of the aluminum. Again, 0.22 kg multiplied by the specific heat capacity of the aluminum 900 joules per kilogram Celsius plus the mass of the water. 9000.15 kg multiplied by the specific heat capacity of the water. 4186 joules per kilogram Celsius. We plug that into our calculator and we are left with 107 degrees Celsius and looking at our multiple choice answers that aligns with answer choice. A. So that's all we have for this one. We'll see you in the next video.

30 g of copper pellets are removed from a 300°C oven and immediately dropped into 100 mL of water at 20°C in an insulated cup. What will the new water temperature be?

Verified Solution

Key Concepts

Heat Transfer

Specific Heat Capacity

Conservation of Energy