Class 11Physics10-1.Thermometry, Thermal Expansion and CalorimetryPrinciple of Calorimetry and Water Equivalent
Easy$A$ glass beaker contains $200 \,g$ of carbonated water initially at $20^{\circ} C$. How much ice should be added to obtain the final temperature of $0^{\circ} C$ with all ice melted, if the initial temperature of ice is $-10^{\circ} C$ (in $\,g$)? Neglect the heat capacity of the glass.
[Take, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]
[Take, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]
- A$47$
- B$76$
- C$200$
- D$22$
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$A$ liquid of mass $m$ and specific heat $c$ is heated to a temperature $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is heated to a temperature $T$. If these two liquids are mixed, the resulting temperature of the mixture is:
Medium
View Solution$1 \,kg$ of ice at $-20^{\circ} C$ is mixed with $2 \,kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment,the final temperature of the mixture is ............ $^{\circ} C$. (Assume,latent heat of ice $= 334.4 \,kJ/kg$,specific heat of water and ice are $4.18 \,kJ \,kg^{-1} K^{-1}$ and $2.09 \,kJ \,kg^{-1} K^{-1}$,respectively.)
MediumKVPY 2015
View SolutionThe time required to raise the temperature of $3 \text{ litre}$ of water from $0^{\circ} C$ to $80^{\circ} C$ by a heater operated under $200 \text{ V}$ having resistance of $50 \Omega$ is
[specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$] [density of water $= 1000 \text{ kg/m}^3$] (in $\text{ min}$)
[specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$] [density of water $= 1000 \text{ kg/m}^3$] (in $\text{ min}$)
$M$ grams of steam at $100^{\circ} C$ is mixed with $200 \; g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 \; cal/g$ and heat of fusion of ice is $80 \; cal/g$],the value of $M$ is:
$1\, g$ of steam at $100^{\circ}C$ melts ........ $g$ of ice at $0^{\circ}C?$ (Latent heat of ice $= 80\, cal/g$ and latent heat of steam $= 540\, cal/g$)
Difficult
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