Chapter 1, Problem 1.66P

1-66 How many calories are required to heat the following (specific heats are given in Table 1-4)?

(a) 52.7 g of aluminum from 100^oC to 285^oC

(b) 93.6 g of methanol from —35^oC to 55^oC

(c) 3.4 kg of lead from —33^oC to 730^oC

(d) 71.4 g of ice from —77^oC to —5^oC

Interpretation Introduction

(a)

Interpretation:

Heat required rise temperature of aluminum from 100 to 285 ° C should be calculated.

Concept Introduction:

Heat required can be calculate using the following formula:

$\text{Q = m s }\Delta \text{T}$

Where, q is heat required,

m is mass of substance in g,

s is specific heat capacity of the substance and,

$\Delta \text{T}$ is change in temperature.

Answer to Problem 1.66P

The heat required is 2145 cal.

Explanation of Solution

Given Information:

The mass of aluminum is 52.7 g and change in temperature takes place from 100 ° C to 285 ° C. The value of specific heat capacity of Al from table 1-4 is $0.22\text{ cal/g}\text{C}$

For aluminum, heat required can be calculated as follows:

$\begin{array}{l}\begin{array}{l}\text{Q = m s }\Delta \text{T}\hfill \\ \text{= }\left(\text{52}\text{.7 g}\right)\left(\text{0}{\text{.22 cal/g }}^{°}\text{C}\right)\left({\text{285 }}^{°}\text{C}-{\text{100 }}^{°}\text{C}\right)\hfill \end{array}\\ \text{= 2145 cal}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

Heat required rise temperature of methanol from − 35 to 55 ° C should be calculated.

Concept Introduction:

Heat required can be calculate using the following formula:

$\text{Q = m s }\Delta \text{T}$

Where, q is heat required,

m is mass of substance in g,

s is specific heat capacity of the substance and,

$\Delta \text{T}$is change in temperature.

Answer to Problem 1.66P

Heat required is 5140 cal.

Explanation of Solution

Given Information:

Mass of methanol is 93.6 g, change in temperature takes place from $-{\text{35}}^{°}\text{C}$to $5{\text{5}}^{°}\text{C}$and specific heat capacity of methanol from table 1-4 is $\text{0}{\text{.61 cal/g }}^{°}\text{C}$

For methanol, heat required can be calculated as follows:

$\begin{array}{l}\text{Q = m s }\Delta \text{T}\hfill \\ \text{= }\left(\text{93}\text{.6 g}\right)\left(\text{ 0}{\text{.61 cal/g }}^{°}\text{C}\right)\left({\text{55 }}^{°}\text{C}-\left(-{\text{35}}^{°}\text{C}\right)\right)\hfill \\ \approx \text{ 5140 cal}\hfill \end{array}$

Interpretation Introduction

(c)

Interpretation:

Heat required to rise the temperature of lead from − 33 to 730 ° C should be calculated.

Concept Introduction:

Heat required can be calculate using the following formula:

$\text{Q = m s }\Delta \text{T}$

Where, q is heat required,

m is mass of substance in g,

s is specific heat capacity of the substance and,

$\Delta \text{T}$is change in temperature.

Answer to Problem 1.66P

Heat required is 80420 cal.

Explanation of Solution

Given Information:

Mass of lead is 3.4 kg, change in temperature takes place from -33 to 730 ° C and specific heat capacity from table 1-4 is 0.031 ${\text{cal/g }}^{°}\text{C}$

For Lead, heat required can be calculated as follows:

$\begin{array}{l}\begin{array}{l}\text{Q = m s }\Delta \text{T}\hfill \\ \text{= 3}\text{.4 kg}\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\times \text{0}{\text{.031 cal/g }}^{°}\text{C}\times \left({\text{730}}^{\circ }\text{C}-\left(-{\text{33 }}^{\circ }\text{C}\right)\right)\hfill \end{array}\\ \text{= 80420 cal}\end{array}$

Interpretation Introduction

(d)

Interpretation:

Heat required to rise the temperature of ice from 100 to 285 ° C should be calculated.

Concept Introduction:

Heat required can be calculate using the following formula:

$\text{Q = m s }\Delta \text{T}$

Where, q is heat required,

m is mass of substance in g,

s is specific heat capacity of the substance and,

$\Delta \text{T}$is change in temperature.

Answer to Problem 1.66P

Heat required is $\text{5243}\text{.6 cal}$.

Explanation of Solution

Given Information:

Mass of ice is 71.4 g, change in temperature takes place from -77 ° C to -5 ° C and specific heat capacity of ice is 0.48 ${\text{cal/g }}^{°}\text{C}$

For ice, heat required can be calculated as follows:

$\begin{array}{l}\text{Q = m s dT}\hfill \\ \text{=}\left(\text{ 71}\text{.4 g}\right)\left(\text{1}{\text{.02 cal/g }}^{°}\text{C}\right)\left(-{\text{5 }}^{°}\text{C}-\left(-{\text{77 }}^{°}\text{C}\right)\right)\hfill \\ \text{= 5243}\text{.6 cal}\hfill \end{array}$