Calorimetry MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\frac{{90 - V}}{{2{R_T}}} = \frac{{V - 30}}{{{R_T}}} \Rightarrow V = {50^ \circ }C$$

$$\eqalign{ & \frac{{\left( {{m_w}{C_w} + W} \right)\Delta T}}{t} = \frac{{\left( {{m_\ell }{C_\ell } + W} \right)\Delta T}}{t}\,\,\left( {W = {\text{water equivalent of the vessel}}} \right) \cr & {\text{or,}}\,\,{m_w}{C_w} = {m_\ell }{C_\ell } \cr} $$
$$\therefore $$ Specific heat of liquid, $$C = \frac{{{m_W}{C_W}}}{{{m_\ell }}}$$
$$ = \frac{{50 \times 1}}{{100}} = 0.5\,kcal/kg$$

$$\eqalign{ & W = JQ \Rightarrow \frac{1}{2}\left( {\frac{1}{2}M{v^2}} \right) = J\left( {m.c.\Delta \theta } \right) \cr & \frac{1}{4} \times 1 \times {\left( {50} \right)^2} = 4.2\left[ {200 \times 0.105 \times \Delta \theta } \right] \cr & \Rightarrow \Delta \theta = {7.1^ \circ }C \cr} $$

Graph (A) shows the variation of temperature with time. At first temperature will increase then there will be state change from liquid to gas.

$$\eqalign{ & Q = mc\,\Delta T \cr & \Rightarrow \,\,Q = mc\left( {T - {t_0}} \right)\,\,\,.....\left( {\text{i}} \right) \cr} $$
∴ From $$50\,K$$  to boiling temperature, $$T$$ increases linearly.
During boiling, equation is
$$Q = mL$$
Temperature remains constant till boiling is complete After that, again eqn. (i) is followed and temperature increases linearly.

According to question as conservation of energy, energy gained by the ice during its fall from height $$h$$ is given by $$E = mgh$$
As given, only one quarter of its energy is absorbed by the ice.
$$\eqalign{ & {\text{So,}}\,\frac{{mgh}}{4} = m{L_f} \Rightarrow h = \frac{{m{L_f} \times 4}}{{mg}} \cr & = \frac{{{L_f} \times 4}}{g} = \frac{{3.4 \times {{10}^5} \times 4}}{{10}} \cr & = 13.6 \times {10^4} = 136000\,m = 136\,km \cr} $$

Initially, on heating temperature rises from $$ - {100^ \circ }C$$  to $$ - {0^ \circ }C.$$  Then ice melts and temperature does not rise. After the whole ice has melted, temperature begins to rise until it reaches $${100^ \circ }C.$$  then it becomes constant, as at the boiling point will not rise.

The heat lost in condensation $$ = x \times 540\,cal.$$
$$\eqalign{ & \therefore x \times 540 = y \times 80 + y \times 1 \times \left( {100 - 0} \right) \cr & {\text{or}}\,\,\frac{x}{y} = \frac{1}{3}. \cr} $$

Rate of heat loss with only room heater
$${P_h} = \frac{{\Delta Q}}{{\Delta t}} = C\left( {16 + 20} \right),C = {\text{constant}}$$
while for both heater and stove it is
$$\eqalign{ & {P_h} + {P_s} = {\left( {\frac{{\Delta Q}}{{\Delta t}}} \right)^\prime } = C\left( {22 + 20} \right) \cr & \frac{{{P_h}}}{{{P_h} + {P_s}}} = \frac{{36}}{{42}} \Rightarrow 7{P_h} = 6{P_h} + 6{P_s} \cr & \Rightarrow {P_h} = 6{P_s} = 6\,kW \cr} $$

Specific heat of water $$= 4200\,J/kg-K$$
Latent heat of fusion $$ = 3.36 \times {10^5}\,J/kg$$
Latent heat of vapourisation
$$\eqalign{ & = 22.68 \times {10^5}\,J/kg \cr & x \times {10^{ - 3}} \times 22.68 \times {10^5}\;J \cr & = y \times {10^{ - 3}} \times 3.36 \times {10^5}\;J + y \times {10^{ - 3}} \times 4200 \times 100 \cr & \therefore \frac{x}{y} = \frac{{7.56}}{{22.68}} = \frac{1}{3} \cr} $$