One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at 0°C to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached 32.0°C. To treat a 70.0-kg patient, what is the minimum amount of ice (at 0°C) you need in the bath so that its temperature remains at 0°C? The specific heat of the human body is 3480 J/kg C°, and recall that normal body temperature is 37.0°C.

Video transcript

Welcome back everybody. We are taking a look at a a slaughterhouse that preserves its meats in ice water combinations. Now, we are looking at a specific package of meat and it starts out with a temperature of 37°C and is cooled down to two degrees Celsius. We're told that the mass of the meat is kg and that the temperature of the ice water mixture used to cool down the meat is zero degrees Celsius. We're also told that the specific heat of the meat is 3480 joules per uh sorry, uh, a j jewels per kilogram degrees Celsius. And we are tasked with finding what amount of ice was the least amount of ice. We need to cool down the 65 kg of meat while still maintaining the temperature of 0°C. Well, if we're maintaining the temperature of zero degrees, we know then that the heat gained by the ice and lost by the meat is going to be equal to our temperature of the ice water bath, which is just going to be zero. This means that the temperature gained by the ice is equal to the negative temperature lost by the meat. So let's go ahead and plug in some variables here. Well, for the temperature gained by the ice, we are going to have the mass of the ice times the latent heat of fusion of the ice, which I'm just going to give by l here and it's a constant, it's gonna be equal to the negative of the mass of the meat, times the specific heat of the meat times the change in temperature of the meat, isolating the mass of the ice. We then have that the mass of the ice is equal to the mass of the meat. Times specific heat of meat times the change in temperature of the meat all over the latent heat of fusion of ice. But we have all of these values. So let's just go ahead and plug in what we know mass of the meat is 65 kg times its specific heat of 3480 times the change in temperature. So you're gonna have your final temperature of two minus the initial temperature of all over. The latent heat of fusion. Which is just given by three 334 times 10 to the third joules per kilogram which when you plug all of this into your calculator, you get 23.7 kg of ice corresponding to our final answer. Choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one