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What Is Heat? Help (page 2)

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By — McGraw-Hill Professional
Updated on Sep 5, 2011

The British Thermal Unit (btu)

In some applications, a completely different unit of heat is used: the British thermal unit (Btu). You’ve heard this unit mentioned in advertisements for furnaces and air conditioners. If someone talks about Btus alone in regard to the heating or cooling capacity of a furnace or air conditioner, this is an improper use of the term. They really mean to quote the rate of energy transfer in Btus per hour, not the total amount of energy transfer in Btus.

The Btu is defined as the amount of heat that will raise or lower the temperature of exactly one pound (1 lb) of pure liquid water by one degree Fahrenheit (1°F). Does something seem flawed about this definition? If you’re uneasy about it, you have a good reason. What is a pound? It depends where you are. How much water weighs 1 lb? On the Earth’s surface, it’s approximately 0.454 kg or 454 g. On Mars, however, it takes about 1.23 kg of liquid water to weigh 1 lb. In a weightless environment, such as on board a space vessel orbiting the Earth or coasting through deep space, the definition of Btu is meaningless because there is no such thing as a pound at all.

Despite these flaws, the Btu is still used once in awhile, so you should be acquainted with it. Specific heat is occasionally specified in Btus per pound per degree Fahrenheit (Btu/lb/°F). In general, this is not the same number, for any given substance, as the specific heat in cal/g/°C.

Heat Practice Problem

Problem

Suppose that you have 3.00 g of a certain substance. You transfer 5.0000 cal of energy to it, and the temperature goes up uniformly throughout the sample by 1.1234°C. It does not boil, condense, freeze, or thaw during this process. What is the specific heat of this stuff?

Solution

Let’s find out how much energy is accepted by 1.00 g of the matter in question. We have 3.00 g of the material, and it gets 5.0000 cai, so we can conclude that each gram gets 1 /3 of this 5.0000 cal, or 1.6667 cal.

We’re told that the temperature rises uniformly throughout the sample. This is to say, it doesn’t heat up more in some places than in other places. It gets hotter to exactly the same extent everywhere. Therefore, 1.00 g of this stuff goes up in temperature by 1.1234°C when 1.6667 cal of energy is transferred to it. How much heat is required to raise the temperature by 1.0000°C? This is the number c we seek, the specific heat. To get c , we must divide 1.6667 cal/g by 1.1234°C. This gives us c = 1.4836 cal/g/°C. Because we are given the mass of the sample to only three significant figures, we must round this off to 1.48 cal/g/°C.

Practice problems of these concepts can be found at: Temperature, Pressure, And Changes Of State Practice Test

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