/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 One suggested treatment for a pe... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The minimum amount of ice required so that its temperature remains at \(0^{\circ} C\) is approximately 3.66 kg.

Step by step solution

01

Determine the heat lost by the patient

First we have to calculate the amount of heat the patient's body will lose when the body temperature drops from \(37.0^{\circ} C\) to \(32.0^{\circ} C\). We use the formula \( q = mc \Delta T \), where \( m = 70.0 kg \) is the mass of the patient, \( c = 3480 J/kgâ‹…^{\circ} C \) is the specific heat of the human body, and \( \Delta T = 37.0^{\circ} C - 32.0^{\circ} C = 5.0^{\circ} C \) is the change in body temperature. So, \( q = 70.0 kg \times 3480 J/kgâ‹…^{\circ} C \times 5.0^{\circ} C = 1.22 \times 10^6 J \)
02

Determine the amount of ice required

Now we need to calculate how much ice is needed to absorb this heat. When ice melts, it absorbs heat from the surroundings, and each gram of ice absorbs 334 J to change state from solid to liquid, still at \(0^{\circ} C\). The mass \( m \) of ice required will be the heat divided by the heat absorbed per gram of ice i.e., \( m = q/L_f \), where \( L_f = 334 J/g \) (latent heat of fusion of ice). So, m = \(1.22 \times 10^6 J / 334 J/g = 3660 g or 3.66 kg\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Specific heat capacity is a property that tells us how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. It basically measures a material's ability to hold heat. In the provided exercise, we talk about the specific heat capacity of the human body, which is given as \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\). Understand this: the higher the specific heat capacity, the more heat is needed to increase the temperature, making the substance a good candidate for thermal treatments like preventing brain damage post-stroke by cooling.

When the patient's body temperature decreases, heat is transferred out of the body and absorbed by the surroundings. This concept is crucial for medical procedures that require precise temperature control, such as the one described where lowering the body temperature is key to the patient's recovery.
Latent Heat of Fusion
Latent heat of fusion is the amount of heat energy required to change the state of a substance from solid to liquid at its melting point, without changing its temperature. It is a crucial concept when dealing with phase changes, like melting ice in our exercise. For ice, the latent heat of fusion is \(334 J/g\), meaning each gram of ice needs to absorb 334 joules to melt.

The exercise mentions using ice to absorb the heat from the patient, taking advantage of the ice's latent heat of fusion. Since the temperature during the phase change remains constant, the ice can absorb a significant amount of heat energy without increasing in temperature, which is exactly what's needed to maintain a steady water temperature in the ice-water bath.
Heat Transfer
Heat transfer involves the movement of thermal energy from one object or material to another. In the realms of medicine and thermal physics, understanding how heat transfer works is fundamental for safely changing a patient's body temperature. There are three mechanisms of heat transfer: conduction, convection, and radiation. In our exercise, the key aspect of heat transfer is through the melting of ice, which is a combination of conduction (direct contact transfer) and the phase change absorption. This heat transfer keeps the water in the bath at a steady \(0^\circ \mathrm{C}\).

Remember, efficient heat transfer mechanisms are vital in medical applications to prevent overheating or undercooling, which can be detrimental to patient health.
Temperature Change in Thermal Systems
In any thermal system, the temperature change is the difference observed in the system's temperature when heat is added or removed. The principle of energy conservation plays a vital role here. As the patient's body temperature decreases, heat is conserved by transferring to the surrounding ice-water bath, maintaining thermal equilibrium.

The exercise illustrates this when the patient's body temperature drops from \(37.0^\circ \mathrm{C}\) to \(32.0^\circ \mathrm{C}\). Monitoring and controlling temperature changes are crucial in medical treatments to prevent tissue damage and, in the case of stroke patients, to limit brain damage by reducing metabolic rates.

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Most popular questions from this chapter

A physicist uses a cylindrical metal can \(0.250 \mathrm{~m}\) high and \(0.090 \mathrm{~m}\) in diameter to store liquid helium at \(4.22 \mathrm{~K} ;\) at that temperature the heat of vaporization of helium is \(2.09 \times 10^{4} \mathrm{~J} / \mathrm{kg} .\) Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, \(77.3 \mathrm{~K},\) with vacuum between the can and walls. How much liquid helium boils away per hour? The emissivity of the metal can is 0.200 . The only heat transfer between the metal can and the surrounding walls is by radiation.

A machinist bores a hole of diameter \(1.35 \mathrm{~cm}\) in a steel plate that is at \(25.0^{\circ} \mathrm{C}\). What is the cross-sectional area of the hole (a) at \(25.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) when the temperature of the plate is increased to \(175^{\circ} \mathrm{C} ?\) Assume that the coefficient of linear expansion remains constant over this temperature range.

A copper pot with a mass of \(0.500 \mathrm{~kg}\) contains \(0.170 \mathrm{~kg}\) of water, and both are at \(20.0^{\circ} \mathrm{C}\). A \(0.250 \mathrm{~kg}\) block of iron at \(85.0^{\circ} \mathrm{C}\) is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

You are making pesto for your pasta and have a cylindrical measuring cup \(10.0 \mathrm{~cm}\) high made of ordinary glass \(\left[\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) that is filled with olive oil \(\left[\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) to a height of \(3.00 \mathrm{~mm}\) below the top of the cup. Initially, the cup and oil are at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\). You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

(a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January \(23,1916 .\) The next day the temperature plummeted to \(-56^{\circ} \mathrm{F}\). What was the temperature change in Celsius degrees?

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