Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

  $\text{heat}=\left(\text{mass}\right)\left(\text{specific}\text{heat}\right)\left(\text{change}\text{in}\text{temeprature}\right)$

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

  $\text{heat lost}=\left(\text{mass of alloy}\right)\left(\text{specific}\text{heat}\right)\left({\text{T}}_2-{\text{T}}_1\right)$        (1)

Here,

  ${\text{T}}_2$ is final temperature.

  ${\text{T}}_1$ is initial temperature.

Substitute $45\text{g}$ for mass of alloy, $37°\text{C}$ for ${\text{T}}_2$ and $100°\text{C}$ for ${\text{T}}_1$ in equation (1).

  $\begin{array}{c}\text{heat lost}=\left(45\text{g}\right)\left(\text{specific}\text{heat}\right)\left(37°\text{C}-100°\text{C}\right)\\ =-2835\text{g}\cdot °\text{C}\left(\text{specific}\text{heat}\right)\end{array}$

The formula to calculate heat gained by water is a follows:

  $\text{heat gained}=\left(\text{mass of water}\right)\left(\text{specific}\text{heat}\right)\left({\text{T}}_2-{\text{T}}_1\right)$        (2)

Substitute $100\text{g}$ for mass of alloy, $4.184\text{J}\cdot {\text{g}}^{-1}\cdot °\text{C}$ for specific heat, $37°\text{C}$ for ${\text{T}}_2$ and $25°\text{C}$ for ${\text{T}}_1$ in equation (2).

  $\begin{array}{c}\text{heat gained}=\left(100\text{g}\right)\left(4.184\text{J}\cdot {\text{g}}^{-1}\cdot °{\text{C}}^{-1}\right)\left(37°\text{C}-25°\text{C}\right)\\ =5020.8\text{J}\end{array}$

Since energy remains conserved so it can be expressed as follows:

  $\text{heat gained}+\text{heat lost}=\text{0}$

Substitute equation $5020.8\text{J}$ for heat gained and $-2835\text{g}\cdot °\text{C}\left(\text{specific}\text{heat}\right)$ for heat lost.

  $\begin{array}{c}5020.8\text{J}+\left(-2835\text{g}\cdot °\text{C}\left(\text{specific}\text{heat}\right)\right)=\text{0}\\ \left(\text{specific}\text{heat}\right)=\frac{-5020.8\text{J}}{-2835\text{g}\cdot °\text{C}}\\ =1.771\text{J}\cdot {\text{g}}^{-1}\cdot °{\text{C}}^{-1}\end{array}$

The specific heat of alloy is $1.771\text{J}\cdot {\text{g}}^{-1}\cdot °{\text{C}}^{-1}$. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is $1.771\text{J}\cdot {\text{g}}^{-1}\cdot °{\text{C}}^{-1}$. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is $1.771\text{J}\cdot {\text{g}}^{-1}\cdot °{\text{C}}^{-1}$. Hence option (B) is correct.