Calorimetry MCQ Questions & Answers in Heat and Thermodynamics | Physics

Concept
Apply principle of calorimetry
According to principle of calorimetry
Heat lost by steam = Heat gained by water
Let $${m'}$$ be the amount of steam that converts into water.
$$\eqalign{ & m' \times L + m's\Delta T = ms\Delta t\,\,\left[ {_{L\, = {\text{ Latent heat of water}}}^{s\, = \,{\text{Specific heat of water}}}} \right] \cr & m' \times 540 + m' \times 1 \times \left( {100 - 80} \right) = 20 \times 1 \times \left( {80 - 10} \right) \cr & m' = \frac{{20 \times 70}}{{560}} = 2.5\,g \cr} $$
Now, net mass of water $$= 20 + 2.5 = 22.5\,g$$

$$\eqalign{ & \frac{C}{5} = \frac{{F - {3^2}}}{9} \Rightarrow \frac{{\Delta C}}{5} = \frac{{\Delta F}}{9} \cr & \Delta C = \frac{5}{9}\Delta F \Rightarrow \Delta C = \Delta K \cr} $$

$$OA$$  refers to change of state from ice to water without change of temperature.

Heat required by $$1\,g$$  ice at $${0^ \circ }C$$  to melt into $$1\,g$$  water at $${0^ \circ }C,$$
$$\eqalign{ & {Q_1} = mL\,\,\left( {L = {\text{latent heat of fusion}}} \right) \cr & = 1 \times 80 = 80\,cal\,\,\left( {L = 80\,cal/g} \right) \cr} $$
Heat required by $$1\,g$$  of water at $${0^ \circ }C$$  to boil at $${100^ \circ }C,$$
$$\eqalign{ & {Q_2} = mc\Delta \theta \,\,\left( {c = {\text{specific}}\,{\text{heat}}\,{\text{of}}\,{\text{water}}} \right) \cr & = 1 \times 1\left( {100 - 0} \right)\,\,\left( {c = 1\,cal/{g^ \circ }C} \right) \cr & = 100\,cal \cr} $$
Thus, total heat required by $$1\,g$$  of ice to reach a temperature of $$10{0^ \circ }C,$$
$$Q = {Q_1} + {Q_2} = 80 + 100 = 180\,cal$$
Heat available with $$1\,g$$  of steam to condense into $$1\,g$$  of water at $${100^ \circ }C,$$
$$\eqalign{ & Q' = mL'\,\,\left( {L' = {\text{latent heat of vaporisation}}} \right) \cr & = 1 \times 536\,cal\,\,\left( {L' = 536\,cal/g} \right) \cr & = 536\,cal \cr} $$
Obviously, the whole steam will not be condensed and ice will attain temperature of $${100^ \circ }C.$$  Thus, the temperature of mixture is $${100^ \circ }C.$$

$$\eqalign{ & {Q_1} = kA\left( {\frac{{800 - 100}}{x}} \right)t = m{L_{{\text{stream}}}} \cr & {Q_2} = kA\left( {\frac{{800 - 0}}{{1 - x}}} \right)t = m{L_{{\text{ice}}}} \cr} $$
Dividing, $$\left( {\frac{{700}}{x}} \right)\left( {\frac{{1 - x}}{{800}}} \right) = 7$$
$${\text{or}}\,1 - x = 8x\,\,{\text{or}}\,\,x = \frac{1}{9}m$$

Heat required to convert $$5\,kg$$  of water at $${20^ \circ }C$$  to $$5\,kg$$  of water at $${0^ \circ }C$$
$$ = m{C_\omega }\Delta T = 5 \times 1 \times 20 = 100\,k\,cal.$$
Heat released by $$2\,kg.$$  Ice at $$ - {20^ \circ }C$$  to convert into $$2\,kg$$  of ice at $${0^ \circ }C$$
$$ = m{C_{{\text{ice}}}}\Delta T = 2 \times 0.5 \times 20 = 20\,k\,cal.$$
How much ice at $${0^ \circ }C$$  will convert into water at $${0^ \circ }C$$  for giving another $$80\,k\,cal$$  of heat
$$\eqalign{ & Q = mL \cr & \Rightarrow \,\,80 = m \times 80 \cr & \Rightarrow \,\,m = 1\,kg \cr} $$
Therefore the amount of water at $${0^ \circ }C$$
$$= 5\,kg + 1\,kg = 6\,kg$$
Thus, at equilibrium, we have, [$$6\,kg$$  water at $${0^ \circ }C + 1\,kg$$   ice at $${0^ \circ }C$$ ].

According to question only one-quarter of the heat produced by falling piece of ice is absorbed in the melting of ice.
i.e., $$\frac{{mgh}}{4} = mL$$
$$ \Rightarrow h = \frac{{4L}}{g} = \frac{{4 \times 3.4 \times {{10}^5}}}{{10}} = 136\,{\text{km}}{\text{.}}$$

Let $$x$$ kg of ice can melt
Using law of Calorimetry,
Heat lost by copper = Heat gained by ice
$$\eqalign{ & 2 \times 400 \times \left( {500 - 0} \right) = x \times 3.5 \times {10^5} \cr & {\text{or}}\,\,x = \frac{{2 \times 400 \times 500}}{{3.5 \times {{10}^5}}} = \frac{8}{7}kg \cr} $$

Rate of loss of heat $$ \propto $$ difference in temperature with the surroundings.
$$\eqalign{ & {\text{At}}\,\,{50^ \circ }C,\frac{{dQ}}{{dt}} = k\left( {50 - 20} \right) = 10,\,\,{\text{where}}\,\,k = {\text{constant}} \cr & \therefore k = \frac{1}{3} \cr} $$
At an angle temperature of $${35^ \circ }C,$$
$$\frac{{dQ}}{{dt}} = \frac{1}{3}\left( {35 - 20} \right)J/s = 5\,J/s$$
Heat lost in 1 minutes
$$ = \frac{{dQ}}{{dt}} \times 60\,J = 5 \times 60\,J = 300\,J = Q$$
Fall in temperature $$ = {0.2^ \circ }C = \Delta \theta .$$
$$Q = c\Delta \theta .$$
Heat capacity $$ = c = \frac{Q}{{d\theta }} = \frac{{300J}}{{{{0.2}^ \circ }C}} = 1500\,J{/^ \circ }C.$$

As $$1\,g$$  of steam at $${100^ \circ }C$$  melts $$8\,g$$  of ice at $${0^ \circ }C.$$
$$10\,g$$  of steam will melt $$8 \times 10\,g$$   of ice at $${0^ \circ }C$$
Water in calorimeter $$ = 500 + 80 + 10g = 590g$$