Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

Thermostat is used in electric apporatas like refrigerator iron etc for automatic cut off Therefore for metallic strips to bend on heating their coefficient of linear expansion should be different.

Initially ice will absorb heat to rise its temprature to $0 ^\circ C $ then its melting takes place. If mi = Initial mass of ice mi-1 = Mass of ice that melts and mw = Initial mass of water By low of mixture Heat gained by ice = Heat lost by water $ \Rightarrow mi \times c \times (20) + mi + L = mwC_w (20) $ $ \Rightarrow 2 \times 0.5 (20) + mi \times 80 = 5 \times 1 \times 20 \Rightarrow mi^{-7} = 1 kg $ So final mass water Initial mass of water + mass of ice that melts = 5 + 1 = 6 kg

If mass of the bullet is mgm then total heat required for bullet to just melt down. $ \theta_1 = mc \triangle \theta + mL $ Now when bullet is stopped bythe obstacle the loss in its mechanical energy = $ {1 \over 2 } cm \times 10 ^ {-3} v^2 9 $ $ (As mg = m \times 10 ^ {-3} kg ) $ As 25% of this energy is absorbed by the $ \theta_2 = { 75 \over 100 } \times {1 \over 2} mv^2 \times 10^{-3} = {3 \over 8} mv^2 \times 10 ^ {-3} J $ Now the bullet melt if $ \theta_2 \geq \theta_1 $

With rise in pressure melting point of ice decreases also ice contracts on melting.

Celsius scale was the first temperature scale and fahrenheit is the smallest unit measuring temperature

If bath assertion and reason are true and the reason is the correct explanation of the assertion.

The potential energy of water molecules is more. The heat given to melt the ice at $ 0 ^\circ C$ used up in increasing the potentical energy of water molecules formed at $ 0 ^\circ C $

Water has maximum density at $ 4 ^\circ C$ on heating above $ 4 ^\circ C $ or cooling below $ 40 ^\circ C $ density of water decreases and its volume increases. Therefore water overflows in both the cases.

$$\begin{gathered} \mathrm{T}= \left({\mathrm{T}}_0+{\mathrm{at}}^2+\mathrm{b} \sin \mathrm{t}\right)\mathrm{K} \\ \frac{\mathrm{dT}}{\mathrm{dt}}= 2\mathrm{at}+\mathrm{b} \cos \mathrm{t} \\ \mathrm{At} \mathrm{t}= \mathrm{\pi } \sec \\ \frac{\mathrm{dT}}{\mathrm{dt}}=2\left(\frac{2}{\mathrm{\pi }}\right)\mathrm{\pi } + \left(-4\right)\mathrm{cos\pi } \\ = 8 \mathrm{K}/\sec \end{gathered}$$

Given specific heat of container,  C= a+bT

The initial temperature of container $T_i$= 127 + 273 = 400 K

and final temperature of container $T_f$= 300 K

${.}^{.}. Heat lost by container= -{}_{T_i}{}^{T_f}\int m_C(a+bT)dT (where m_C\to mass of container)$

Heat lost= $-{}_{400}{}^{300}\int m_C(a+bT)dT$

= $-m_C{\left[aT+\frac{b{T}^2}{2}\right]}_{400}^{300}$

$$\begin{gathered} = -m_C\left[(a300-a400)+\frac{b}{2}\left(\right(300{)}^2-\left(400{)}^2\right)\right] \\ = m_C\left[100a+35000b\right] \\ = m_C\left[100\times 0.1\times {10}^3+35000\times 40\times {10}^{-3}\right] \\ = m_C\left[10000+1400\right] =m_{C}11400 \end{gathered}$$

Heat gained by ice = $m_{ice}{L}_F+m_{ice}{C}_{water}∆T$

=$(0.1\times 80\times {10}^3)+(0.1\times {10}^3\times 27) =8000+2700$

= 10700 cal

From principle of calorimetry

Heat lost by container= Heat gained by ice

${.}^{.}.$ $$\begin{gathered} 11400m_C=10700 \\ {m}_C=\frac{107}{114}=0.939 kg =939 g \end{gathered}$$