A calorimeter of mass m contains an equal mass of water in it- The temperature of the water and calorimeter is t 2-0- C- A block of ice of mass m and temperature t30B- C 1 t 2- C 2 t 2- L - C 3 t 3-0C- C 1 t 2- C 2 t 2- L - C 3 t 3-0D- None of these

The correct option is C C_1 t_2+ C_2 t_2+L C_3 t_3 lt;0If whole mixture becomes ice, first calorimeter will reach 0^∘C. So, heat given = m C_1 t_2+ mC_2 t_ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Calorimetry

A calorimeter...

Question

A calorimeter of mass m contains an equal mass of water in it. The temperature of the water and calorimeter is $t_{2} > 0^{\circ} C$ . A block of ice of mass m and temperature $t_{3} < 0^{\circ} C$ is gently dropped into the calorimeter. Let $C_{1}, C_{2}$ and $C_{3}$ be the specific heats of material of calorimeter, water and ice respectively and L be the Latent heat of ice. The whole mixture in the calorimeter becomes ice if

C_{1} t_{1} + C_{2} t_{2} + L + C_{3} t_{3} > 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{1} t_{2} + C_{2} t_{2} + L + C_{3} t_{3} < 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{1} t_{2} + C_{2} t_{2} + L - C_{3} t_{3} < 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $C_{1} t_{2} + C_{2} t_{2} + L - C_{3} t_{3} < 0$
If whole mixture becomes ice, first calorimeter will reach 0^\circ C. So, heat given = $m C_{1} t_{2} + m C_{2} t_{2}$ . It is used by ice to increase its temperature

Suggest Corrections

Similar questions

Calculate the heat of fusion of ice from the following data for ice at $0^{\circ} C$ added to water. Mass of calorimeter = 60gm, mass of calorimeter + water = 460 gm, mass of calorimeter + water + ice = 618 gm, initial temperature of water = $38^{\circ} C$ , final temperature of the mixture = $5^{\circ} C$ . The specific heat of calorimeter = $0.10 c a l / g /^{\circ} C$ . Assume that the calorimeter was also at $0^{\circ} C$ initially.

5 g m

of water at

30^{\circ} C

and

5 g m

of ice at

- 20^{\circ} C

are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice

= 0.5 c a l / g m^{\circ} C

and latent heat of ice

= 80 c a l / g m

Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed, If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is

Three liquids with masses $m_{1}, m_{2}, m_{3}$ are thoroughly mixed. If their specific heats are $c_{1}, c_{2}, c_{3}$ and their temperatures are $T_{1}, T_{2}, T_{3}$ respectively, then the temperature of the mixture is: