Chapter 4, Problem 4.61AP

(a)

Interpretation Introduction

Interpretation:

The energy required to heat $\text{50}\text{g}$ of water has to be given in calories and joules.

Concept introduction:

The specific heat is the amount of heat energy needed to raise the temperature of $\text{1g}$ of a substance by ${\text{1}}^{\text{o}}\text{C}$.

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

Answer to Problem 4.61AP

The energy require to heat in calories is $1750\text{cal}\text{.}$

The energy require to heat in Joules is $7315\text{J}\text{.}$

Explanation of Solution

Given,

The mass of water is $\text{50}\text{g}\text{.}$

The temperature at initial $\left({\text{T}}_{\text{1}}\right)$ is ${\text{15}}^{\text{o}}\text{C}\text{.}$

The temperature at final $\left({\text{T}}_{\text{2}}\right)$ is ${\text{50}}^{\text{o}}\text{C}\text{.}$

The specific heat of water in terms of calories is $1.00\text{cal/}{\text{g}}^{\text{o}}\text{C}\text{.}$

The specific heat of water in terms of Joules is $4.18{\text{J/g}}^{\text{o}}\text{C}\text{.}$

The change in temperature $\left(\text{ΔT}\right)$ can be calculated as,

  ${\text{ΔT=T}}_{\text{2}}{\text{-T}}_{\text{1}}$

  $\text{ΔT}{\text{=50}}^{\text{o}}{\text{C-15}}^{\text{o}}\text{C}$

  $\text{ΔT}\text{=}{\text{35}}^{\text{o}}\text{C}$

The energy require to heat in calories can be calculated as,

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

  $\text{Heat=Specific}\text{heat}\times \text{mass×ΔT}$

  $\text{Heat=1}\text{.00}{\text{cal/g}}^{\text{o}}\text{C×50}{\text{g×35}}^{\text{o}}\text{C}$

  $\text{Heat=}1750\text{cal}$

The energy require to heat in calories is $1750\text{cal}\text{.}$

The energy require to heat in Joules can be calculated as,

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

  $\text{Heat=Specific}\text{heat}\times \text{mass×ΔT}$

  $\text{Heat=4}\text{.18}{\text{J/g}}^{\text{o}}\text{C×50}{\text{g×35}}^{\text{o}}\text{C}$

  $\text{Heat=}7315\text{J}$

The energy require to heat in Joules is $7315\text{J}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The energy released when $\text{250}\text{g}$ of aluminium is cooled has to be given in calories and joules.

Concept introduction:

Refer to part (a).

Answer to Problem 4.61AP

The energy released in calories is $-4012.5\text{cal}\text{.}$

The energy released in Joules is $\text{-16781}\text{.25}\text{J}\text{.}$

Explanation of Solution

Given,

The mass of aluminium is $\text{250}\text{g}\text{.}$

The temperature at initial $\left({\text{T}}_{\text{1}}\right)$ is ${\text{125}}^{\text{o}}\text{C}\text{.}$

The temperature at final $\left({\text{T}}_{\text{2}}\right)$ is ${\text{50}}^{\text{o}}\text{C}\text{.}$

The specific heat of aluminium in terms of calories is $\text{0}\text{.214}\text{cal/}{\text{g}}^{\text{o}}\text{C}\text{.}$

The specific heat of aluminium in terms of Joules is $\text{0}\text{.895}{\text{J/g}}^{\text{o}}\text{C}\text{.}$

The change in temperature $\left(\text{ΔT}\right)$ can be calculated as,

  ${\text{Δt=T}}_{\text{2}}{\text{-T}}_{\text{1}}$

  $\text{Δt}{\text{=50}}^{\text{o}}{\text{C-125}}^{\text{o}}\text{C}$

  $\text{Δt}\text{=}-{75}^{\text{o}}\text{C}$

The energy released in calories can be calculated as,

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

  $\text{Heat=Specific}\text{heat}\times \text{mass×ΔT}$

  $\text{Heat=0}\text{.214}{\text{cal/g}}^{\text{o}}\text{C×250}{\text{g×-75}}^{\text{o}}\text{C}$

  $\text{Heat=}-4012.5\text{cal}$

The energy released in calories is $-4012.5\text{cal}\text{.}$

The energy released in Joules can be calculated as,

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

  $\text{Heat=Specific}\text{heat}\times \text{mass×ΔT}$

  $\text{Heat=0}\text{.895}{\text{J/g}}^{\text{o}}\text{C×250}{\text{g×-75}}^{\text{o}}\text{C}$

  $\text{Heat=}-\text{16781}\text{.25}\text{J}$

The energy released in Joules is $\text{-16781}\text{.25}\text{J}\text{.}$