Question

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What quantity of heat is transferred when a 150.0 grams block of iron metal is heated from 25.0 degree Celsius to 75 degree Celsius ?

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Answer 1

Answer:

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Explanation:

Heat Release Rate (HRR) is the rate at which fire releases energy - this is also known as power. HRR is measured in units of Watts (W), which is an International System unit equal to one Joule per second.

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Answer 2

Heat is a familiar manifestation of energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand, so we can conclude that differences in temperatures are the ultimate cause of heat transfer.

Suppose we consider the transfer of heat from the opposite perspective—namely, what happens to a system that gains or loses heat? Generally, the system’s temperature changes. (We will address a few exceptions later.) The greater the original temperature difference, the greater the transfer of heat, and the greater the ultimate temperature change. The relationship between the amount of heat transferred and the temperature change can be written as

heat∝ΔT(7.3.1)(7.3.1)heat∝ΔT

where ∝ means “is proportional to” and ΔT is the change in temperature of the system. Any change in a variable is always defined as “the final value minus the initial value” of the variable, so ΔT is Tfinal − Tinitial. In addition, the greater the mass of an object, the more heat is needed to change its temperature. We can include a variable representing mass (m) to the proportionality as follows:

heat∝mΔT(7.3.2)(7.3.2)heat∝mΔT

To change this proportionality into an equality, we include a proportionality constant. The proportionality constant is called the specific heat and is commonly symbolized by cc:

heat=mcΔT(7.3.3)(7.3.3)heat=mcΔT

Every substance has a characteristic specific heat, which is reported in units of cal/g•°C or cal/g•K, depending on the units used to express ΔT. The specific heat of a substance is the amount of energy that must be transferred to or from 1 g of that substance to change its temperature by 1°. Table 7.3.17.3.1 lists the specific heats for various materials.

Table 7.3.17.3.1: Specific Heats of Selected Substancesaluminum (Al)0.215aluminum oxide (Al2O3)0.305benzene (C6H6)0.251copper (Cu)0.092ethanol (C2H6O)0.578hexane (C6H14)0.394hydrogen (H2)3.419ice [H2O(s)]0.492iron (Fe)0.108iron(III) oxide (Fe2O3)0.156mercury (Hg)0.033oxygen (O2)0.219sodium chloride (NaCl)0.207steam [H2O(g)]0.488water [H2O(ℓ)]1.00
The direction of heat flow is not shown in heat = mcΔT. If energy goes into an object, the total energy of the object increases, and the values of heat ΔT are positive. If energy is coming out of an object, the total energy of the object decreases, and the values of heat and ΔT are negative.

Example 7.3.17.3.1

What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow?

Solution

We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the iron is 73.3°C and the initial temperature is 25.0°C, ΔT is as follows:

ΔT=Tfinal−Tinitial=73.3oC−25.0oC=48.3oCΔT=Tfinal−Tinitial=73.3oC−25.0oC=48.3oC

The mass is given as 150.0 g, and Table 7.3.17.3.1 gives the specific heat of iron as 0.108 cal/g•°C. Substitute the known values into heat = mcΔT and solve for amount of heat:

heat=(150.0g)(0.108calg⋅∘C)(48.3∘C)=782calheat=(150.0g)(0.108calg⋅∘C)(48.3∘C)=782cal

Note how the gram and °C units cancel algebraically, leaving only the calorie unit, which is a unit of heat. Because the temperature of the iron increases, energy (as heat) must be flowing into the metal.

Exercise 7.3.17.3.1

What quantity of heat is transferred when a 295.5 g block of aluminum metal is cooled from 128.0°C to 22.5°C? What is the direction of heat flow?

Example 7.3.27.3.2

A 10.3 g sample of a reddish-brown metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. What is the specific heat of the metal? Can you identify the metal from the data in Table 7.3.17.3.1?

Solution

The question gives us the heat, the final and initial temperatures, and the mass of the sample. The value of ΔT is as follows:

ΔT = Tfinal − Tinitial = 22.0°C − 97.5°C = −75.5°C

If the sample gives off 71.7 cal, it loses energy (as heat), so the value of heat is written as a negative number, −71.7 cal. Substitute the known values into heat = mcΔT and solve for c:

−71.7 cal = (10.3 g)(c)(−75.5°C)

c=−71.7cal(10.3g)(−75.5∘C)c=−71.7cal(10.3g)(−75.5∘C)

c = 0.0923 cal/g•°C

This value for specific heat is very close to that given for copper in Table 7.3.17.3.1.

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