The Physics of Everyday Phenomena
The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
Question
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Chapter 10, Problem 1SP

(a)

To determine

The temperature change of the object due to the addition of given amount of heat, in units of degree Fahrenheit.

(a)

Expert Solution
Check Mark

Answer to Problem 1SP

The temperature change of the object due to the addition of heat, in units of degree Fahrenheit is 90°F.

Explanation of Solution

Given info: The initial temperature of the object is 30°C and its final temperature is 80°C.

Write the expression to find the change in temperature of the object.

ΔT=T2T1 (1)

Here,

T1 is the initial temperature

T2 is the final temperature

Write the formula for converting temperature in degree Celsius to degree Fahrenheit.

T(F)=95T(°C)+32 (2)

Here,

T(F) is the temperature in degree Fahrenheit

T(°C) is the temperature in degree Celsius

Use equation (2) in (1).

ΔT(°F)=(95T2(°C)+32)(95T1(°C)+32)=95T2(°C)95T1(°C)=95(T2(°C)T1(°C)) (3)

Substitute 80°C for T2 and 30°C for T1 in equation (3) to find the change in temperature of the object in units of degree Fahrenheit.

ΔT(°F)=95(80°C30°C)=90°F

Conclusion:

Therefore, the temperature change of the object in degree Fahrenheit is 90°F.

(b)

To determine

The temperature change of the object in Kelvin scale.

(b)

Expert Solution
Check Mark

Answer to Problem 1SP

The temperature change of the object in Kelvin scale is 50K.

Explanation of Solution

The change in temperature of the object in Celsius scale is obtained as 50°C.

Write the formula for converting the temperature in Celsius scale to Kelvin scale.

T(K)=T(°C)+273 (4)

Here,

T(K) is the temperature in Kelvin scale.

Use equation (4) in (1).

ΔT(K)=(T2(°C)+273)(T1(°C)+273)=T2(°C)T1(°C) (5)

Substitute 80°C for T2 and 30°C for T1 to find the change in temperature in the units of Kelvin.

ΔT(K)=80°C30°C=50K

Conclusion:

Therefore, The temperature change of the object in Kelvin scale is 50K.

(c)

To determine

Whether there is any difference in the numerical value of heat capacity expressed in cal/g°C from the one expressed in cal/gK.

(c)

Expert Solution
Check Mark

Answer to Problem 1SP

There is no difference in the numerical value of heat capacity expressed in cal/g°C from the one expressed in cal/gK.

Explanation of Solution

The specific heat of a substance is the amount of heat energy required to raise the temperature of a substance by 1°C or 1K. It is the characteristic of a particular material.

Write the expression for the temperature difference in the units of Kelvin scale.

ΔT(K)=T2(K)T1(K) (6)

Use T(K)=T(°C)+273 in the right hand side of equation (6).

ΔT(K)=(T2(°C)+273)(T1(°C)+273)=T2(°C)T1(°C)=ΔT(°C)

Thus, the magnitude of change in temperature is same irrespective of the measurement in Kelvin scale or Celsius scale. This implies that the increase of 1°C is equivalent to the increase of 1K.

Therefore, the heat energy required to raise the temperature of a substance by 1°C will be equal to the raise in temperature by 1K. In other words, the magnitude of specific heat will be same when it is written in units of cal/g°C or cal/gK.

Conclusion:

Therefore, there is no difference in the numerical value of heat capacity expressed in cal/gC° from the one expressed in cal/gK.

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Chapter 10 Solutions

The Physics of Everyday Phenomena

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