Chapter 10, Problem 65A

Solution

(a)

Interpretation:

The amount of heat absorbed by the water is to be calculated.

Concept introduction:

The amount of heat absorbed by the water can be calculated from the equation;

  $\text{Q}\text{=}\text{m}\text{c}\text{ΔT}$

Where,

Q= amount of heat lost or gained by the sample

m = mass of the sample (in g /kg)

c = specific heat capacity of sample (in J/g ^oC or J/kg ^oC)

  $\text{ΔT}$ = Temperature change (in ^oC)

The amount of heat that is absorbed by water is 1460.2 J

The quantity of heat given to a body, Q can be calculated using the formula,

  $\text{Q}\text{=}\text{m}\text{c}\text{ΔT}$

Given, mass of water m = 100.0 g

Specific heat capacity of water, c = 4.184 J/g ⁰C

Temperature change, $\text{ΔT}$ (Final temperature − initial temperature)

= ${24.8}^{o}C-{21.31}^{o}C$

  $={3.49}^{o}C$

On substituting these values in the above equation,

  $\text{Q}\text{=}\text{m}\text{c}\text{ΔT}$

  $\begin{array}{l}Q=100g\times 4.184J/g^{o}C\times 3.49{}^{o}C\\ =1460.21J\end{array}$

(b)

Interpretation

Heat absorbed by the water has to be compared to the heat lost by the metal.

Concept introduction

An assumption can be made of perfect heat transfer that the heat lost by the metal winds up in the water.

The amount of heat lost by the metal = the amount of heat absorbed by the water

  $Q_{metal}=Q_{water}$

Initially, metal is at higher temperature as it is kept in boiling water. Upon placing the metal in low temperature water, heat will flow from higher temperature metal to the water.

As a result the temperature of the metal will decrease and the temperature of water will increase until the two substances have the same temperature, that is when they reach thermal equilibrium. This assumes that no heat is lost from the water to its surroundings.

(c)

Interpretation: The initial temperature of ‘hot’ metal and the final temperature of the metal is to be calculated.

Concept introduction

The temperature of the metal increases as it is kept in the boiling water. It happens when a known amount of a substance is heated at a temperature and mixed with known amount of water at a given temperature.

• The initial temperature of the ‘hot’ metal is 100^oC.
• The final temperature of the metal is 24.8^oC.

Since the metal gains temperature by keeping in boiling water, it attains the temperature of boiling water, thus making the initial temperature to be 100^oC.

Heat lost by the metal = Heat gained by the water

The water which was at low temperature (given 21.31^oC) gets raised in temperature on keeping the hot metal in the water. Thus, the final temperature of metal becomes 24.8^oC.

(d)

Interpretation:

The specific heat capacity of the metal is to be calculated.

Concept introduction:

Specific heat capacity is the amount of heat needed to raise the temperature of one kilogram of a substance by one Celsius degree.

Specific heat capacity is an intensive property, which doesn’t depend on the quantity of the matter. For water the value of specific heat capacity is 1 Calorie which is equivalent to 4.184 Joules.

The specific heat capacity of the metal is 0.388 J/ g ^oC.

The quantity of heat is Q.

The change in temperature due heat and mass is represented as $\Delta T$

For mass, m the heat is given as Q thus, for 1 kg of mass, heat will be Q/m.

Also, for $\Delta T$ the heat is given as Q/m thus, for raise in 1^oC, amount of heat will be:

  $\frac{Q}{m}\times \Delta T$

Now,

  $\begin{array}{l}Heatlostbythemetal=Heatabsorbedbythewater\\ Q_{metal}=Q_{water}\\ {(}_m={(}_m\end{array}$

Thus,

  $\begin{array}{l}\Delta T_{metal}=Finaltemperature-initialtemperature\\ ={100}^{o}C-{24.8}^{o}C\\ ={75.2}^{o}C\end{array}$

Or,

  $\begin{array}{l}\Delta T_{water}=Finaltemperature-initialtemperature\\ ={24.8}^{o}C-{21.31}^{o}C\\ ={3.49}^{o}C\end{array}$

  ${(mc\Delta T)}_{metal}={(mc\Delta T)}_{water}$

Mass of the metal given = 50.0 g

  $\begin{array}{l}50g\times c\times ({100}^{o}C-{24.8}^{o}C)=100g\times 4.184J/g^{o}C\times ({24.8}^{o}C-{21.31}^{o}C)\\ c\times 3760g^{o}C=1460.216J\\ c=\frac{1460.216J}{3760g^{o}C}\\ =0.388J/g^{o}C\end{array}$

In the above equation, absolute value of ?T is used. If not, then Q_metal = - Q _water

(e)

Interpretation:

The identity of the metal is to be determined.

Concept Introduction:

One can identify a metal from the value of specific heat capacity. It is a physical property of the material a substance is composed of and can be used to help identify the substance.

In comparison with the specific heat capacities, the given metal is found out to be Zinc.

The heat lost by the metal as it cooled is same amount absorbed by the water. Using this principle the specific heat capacity of an unknown metal can be found out.