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This is second of our many quizzes on Specific heat capacity. Click on this to access more such competitive quizzes on Thermal Physics.

Specific heat capacity Quiz 2

physics

A barista pours $24\,g$ of whole milk at room temperature into a cup of $12\,g$ of brewed espresso to make a latte. If the specific heat capacity of the milk is $3.9\,Jg^{-1}{}^{\circ}C^{-1}$ and the resultant specific heat capacity of the latte is $3.07\,Jg^{-1}{}^{\circ}C^{-1}$ then determine the specific heat capacity of the espresso $(\text{in}\,Jg^{-1}{}^{\circ}C^{-1}).$ 

Determine how the specific heat capacity of a mixture can be derived accounting for the mass-specific individual heat capacities.  

  $\text{Given that the mass of the whole milk is}\,m_1 = 24\,g\,\text{ and mass of espresso is}\,12\,g.\\ \text{Therefore, the mass of the mixture/latte will be}\,m = m_1+m_2 = 24+12 = 36\,g\\ \text{Now, given that the specific heat capacity of milk is}\,s_1 = 3.9\,Jg^{-1}\,^{\circ}C^{-1} \\ \text{and that of the mixture is}\,s = 3.07\,Jg^{-1}\,^{\circ}C^{-1},\\ \text{We can find the specific heat capacity of the espresso as follows.}\\ \text{We know that for the given mixture:}\\ s = \left(\dfrac{m_1}{m}\right)s_1 + \left(\dfrac{m_2}{m}\right)s_2\\ \Rightarrow 3.07 = \left(\dfrac{24}{36}\right)3.9 + \left(\dfrac{12}{36}\right)s_2\\ \Rightarrow 3.07 = 2.6 + \dfrac{s_2}{3}\\ \therefore s_2 = (3.07-2.6) \times 3 = 1.41$

The region of space beyond the confines of our atmosphere is an extremely harsh environment with harmful radiation emanating from the sun as well as galactic cosmic rays, both of which majorly consist of energetic protons and neutrons.  Lithium hydride is very effective in bolstering space radiation through relatively shorter penetrative distances without heating up much. If $1\,TeV$ of radiation is incident on $0.34\,g$ of $LiH$ then determine the change in temperature of $LiH$ upon absorption of this radiation. $(\text{specific heat capacity of}\,LiH\,\text{is}\,3.51\,Jg^{-1}K^{-1})$ 

Lithium hydride gains energy from the radiation that it absorbs. Find this energy that causes any change in its temperature.  

 $\text{We are given that}\,LiH\,\text{absorbs all incident radiation, i.e.,}\\ \Delta Q=1\,TeV\\ \Rightarrow\Delta Q = 1 \times 10^{12} \times 1.6 \times 10^{-19}\,J\\ \Rightarrow\Delta Q = 1.6 \times 10^{-7}\,J\\ \text{Given that}\,m = 0.34\,g\,\text{and}\,s = 3.51\,Jg^{-1}K^{-1}\,\text{the change in temperature of}\,LiH\,\text{can be given as:}\\ \Delta T = \dfrac{\Delta Q}{ms}\\ \Rightarrow\Delta T = \dfrac{1.6 \times 10^{-17}}{0.34 \times 3.51}\\ \therefore\Delta T = 1.34 \times 10^{-7}\,K$

Fill in the blanks: You decide to come up with a way to measure the specific heat capacity of a solid of mass $m=500\,g$ as shown in the figure. The solid is first placed in a trough filled with ice cold water and left until the solid is at the same temperature as the water and no more heat transfer occurs, say at $0.5^{\circ}C$. Now, you take $250\,g$ of water at $80^{\circ}C$ in another trough and transfer the solid to this trough until there is no heat transfer between the water and the solid, say at $74^{\circ}C$. The energy transferred from the warm water to the solid is ______$\,J$ and hence, the specific heat capacity of the solid is found to be ________$\,Jkg^{-1}{}^{\circ}C^{-1}.$$(\text{Take the specific heat capacity of water is}\,s_w = 4180\,Jkg^{-1}{}^{\circ}C^{-1})$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/710fb7e6-ba4b-4499-bfdb-6571ec99e6675246290698479287533.png"/> 

Find the amount of heat lost by the warm water which will be equivalent to the amount of heat gained by the solid, from which you can arrive at its specific heat.  

$\text{Given that the mass of the solid is}\,m = 500\,g = 0.5\,kg.\\ \text{When the solid is transferred into the trough with warm water:}\\ \text{The mass of warm water is}\,m_{w} = 250\,g = 0.25\,kg.\\ \text{The specific heat capacity of water is}\,s_w = 4180\,Jkg^{-1}{}^{\circ}C^{-1}\\ \text{The change in temperature of the water is}\,\Delta T = 80-74 = 6^{\circ}C\\ \text{The heat transferred lost by the warm water can be given as:}\\ \Delta Q = m_w s_{w} \Delta T\\ \Rightarrow\Delta Q = 0.25 \times 4180 \times 6\\ \Rightarrow\Delta Q = 6270\,J\\ \text{The heat lost by the water is equivalent to the amount of heat gained by the solid,}\\ \text{which changes the temperature of the solid through:}\,\Delta T = 74-0.5 = 73.5^{\circ}C\\ \text{If s is the specific heat capacity of the solid then,}\\ \Delta Q = ms\Delta T\\ \Rightarrow 6270 = 0.5 \times s \times 73.5\\ \Rightarrow s = \dfrac{6270}{0.5 \times 73.5}\\ \therefore s = 170.6\,Jkg^{-1}\,^{\circ}C^{-1}\\ \text{Therefore, in the context of the question:}\\\text{“…The energy transferred from the warm water to the solid is 6270 J}\\ \text{and hence, the specific heat capacity of the solid is found to be}\,170.6\,Jkg^{-1}{}^{\circ}C^{-1}.”$

 A meteorological balloon is a high-altitude balloon that usually contains instruments capable of measuring atmospheric conditions specific to the altitude in the atmosphere where the readings are taken. These balloons are usually two types: hydrogen based and helium based. Hydrogen is a low-cost gas that is lightest in comparison to air and is used in balloons that record higher altitude readings whereas helium is used in those balloons that are at relatively lower altitudes where safety is of utmost priority.$\\$ Assuming that these balloons are fully expandable, requiring zero energy expenditure to carry out any expansions in volume, determine the ratio of heat required to raise the same amount of gas by the same temperature in both cases respectively.  

Determine the specific heat capacity of hydrogen and helium at constant pressure since there is no restriction to expansion and arrive at the ratio of their heat expended. 

 $\text{Since the balloons are completely expandable,}\\ \text{this means that any expansion is brought about under constant pressure.}\\ \text{The heat involved in the processes will thus be concerned only}\\ \text{with the specific heat of the gas at constant pressure}\,C_P.\\ \text{Additionally, we have the same amounts of gas in each case,}\\ \text{leading to same changes in temperature, therefore, from,}\\ \Delta Q = nC_P\Delta T\,\text{we have:}\\ \Delta Q \propto C_P\\ \text{For the hydrogen}\, H_2\,\text{balloon}:(\text{diatomic gas})\\ \text{Degrees of freedom}\,f = 5\\ C_V = \dfrac{f}{2}R\\ \Rightarrow C_V = \dfrac{5}{2}R\\ \Rightarrow C_P = C_V+R\\ \Rightarrow C_P  = \dfrac{5}{2}R+R\\ \Rightarrow C_P  = \dfrac{7}{2}R\\ \text{Therefore}\,Q_{H_{2}} \propto \dfrac{7}{2}R.\\ \text{Similarly for helium}\,He\,\text{balloon}: (\text{monoatomic gas})\\ \text{Degrees of freedom f = 3}\\ C_V = \dfrac{f}{2}R\\ \Rightarrow C_V  = \dfrac{3}{2}R\\ \Rightarrow C_P = C_V +R\\ \Rightarrow C_P = \dfrac{3}{2}R +R\\ \Rightarrow C_P = \dfrac{5}{2}R\\ \text{Therefore}\, Q_{He} \propto \dfrac{5}{2}R.\\ \text{The ratio of the heats will be:}\\ \dfrac{Q_{H_2}}{Q_{He}} = \dfrac{7R}{5R} \\ \Rightarrow \dfrac{Q_{H_2}}{Q_{He}} = \dfrac{7}{5}\\ \therefore Q_{H_2}: Q_{He} = 7:5$

The Ozone Layer is a region in the Earth’s stratosphere that contains relatively high concentrations of ozone gas and is responsible for absorbing most of the Sun’s harmful UV radiation. If harmful radiation of wavelength $280\,nm$ is incident per $10^{-20}$ mole of ozone then determine the change in temperature of the ozone molecules upon complete absorption of this radiation provided that gas pressure remains constant throughout. $(\text{Take}\,R = 8.31\,J\,mol^{-1}K^{-1})$

First find the energy associated with the UV photon and then find the specific heat capacity of ozone gas following which you can arrive at the change in temperature.   

$\text{We are given that the wavelength of harmful radiation is,}\\ \lambda = 280\,nm = 280 \times 10^{-9}\,m\\ \text{The energy associated with these UV photons will be:}\\ E = \dfrac{hc}{\lambda}\\ \Rightarrow  E = \dfrac{(6.626 \times 10^{-34}) \times (3 \times 10^{8})}{280 \times 10^{-9}}\\ \Rightarrow  E = 0.071 \times 10^{-17}\,J\\ \text{Now, ozone is a non-linear triatomic gas and hence has f=6 degrees of freedom.}\\ C_V = \dfrac{f}{2}R\\ \Rightarrow C_V = \dfrac{6}{2}R \\ \Rightarrow C_V = 3R \\ \text{Since we are given that the pressure remains constant throughout,} \\ C_P = C_V +R \\ \Rightarrow C_P = 3R+R = 4R \\ \Rightarrow C_P = 4 \times 8.31 \\ \Rightarrow C_P = 33.24\,J\,mol^{-1}K^{-1} \\ \text{The ozone molecules completely absorb the incident UV photons, i.e.,}\\ E = \Delta Q\\ 0.071 \times 10^{-17} = nC_P\Delta T \\ \Rightarrow \Delta T = \dfrac{0.071 \times 10^{-17}}{nC_P} \\\Rightarrow \Delta T = \dfrac{0.071 \times 10^{-17}}{10^{-20} \times 33.24} \\ \therefore \Delta T= 2.14\,K$