Home
Class 11
PHYSICS
10 gm of ice at -20^(@)C is dropped into...

`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

A

20 g of water

B

20 g of ice

C

10g ice and 10 g water

D

5 g ice and 15 g water

Text Solution

Verified by Experts

The correct Answer is:
C
Heat given of water `(Q_1) = m_w s_w Delta T`
`= 10 xx 1 xx 10 = 100` cal
Heat absorbed by ice `(Q_2) = mL + m_i s_i Delta T`
`= 10 xx 80 + 10 xx 0.5 xx 20`
`= 800 + 100 = 900 cal`
Since, `Q_1 lt Q_2` (Ice will not completely melt)
Heat supplied by water is only just sufficient to raise the temperature of ice from `-20^@ C` to `0^@ C`, so mixture will be of g 10 ice and g 10 water
Promotional Banner

Similar Questions

Explore conceptually related problems

5 g of ice at 0^(@)C is dropped in a beaker containing 20 g of water at 40^(@)C . The final temperature will be

50 g ice at 0^(@)C is dropped into a calorimeter containing 100 g water at 30^(@)C . If thermal . capacity of calorimeter is zero then amount of ice left in the mixture at equilibrium is

A piece of metal of mass 112 g is heated to 100^@ C and dropped into a copper calorimeter of mass 40 g containing 200 g of water at 16^@ C . Neglecting heat loss, the specific heat of the metal is nearly, if the equilibrium temperature reached is 24^@ C . (S_(cu) = 0.1 cal//g-.^(@) C) .

500 gm of ice at – 5^(@)C is mixed with 100 gm of water at 20^(@)C when equilibrium is reached, amount of ice in the mixture is:

How many gm of ice at -20^(@)C are needed to cool 200 gm of water from 25^(@)C to 10^(@)C ?

10 gm of ice at – 20^(@)C is added to 10 gm of water at 50^(@)C . Specific heat of water = 1 cal//g–.^(@)C , specific heat of ice = 0.5 cal//gm-.^(@)C . Latent heat of ice = 80 cal/gm. Then resulting temperature is -

Four cubes of ice at -10^(@)C each one gm is taken out from the refrigerator and are put in 150 gm of water at 20^(@)C . The temperature of water when thermal equilibrium is attained. Assuming that no heat is lost to the outside and water equivalent of contaner is 46 gm . (Specific heat capacity of water = 1 cal//gm-^(@)C , Specific heat capacity of ice =0.5 cal//gm-^(@)C , Latent heat of fusion of ice = 80 cal//gm-^(@)C )

A calorimeter contains 10 g of water at 20^(@)C . The temperature falls to 15^(@)C in 10 min. When calorimeter contains 20 g of water at 20^(@)C , it takes 15 min for the temperature to becomes 15^(@)C . The water equivalent of the calorimeter is

A calorimeter of mass m contains an equal mass of water in it. The temperature of the water and clorimeter is t_(2).A block of ice of mass m and temperature t_(3)lt0^(@)C is gently dropped into the calorimeter. Let C_(1),C_(2) and C_(3) be the specific heats of calorimeter, water and ice respectively and L be the latent heat of ice. Water equivalent of calorimeter is

An electrically heated coil is immersed in a calorimeter containing 360 g of water at 10^@C . The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and coil is 40 g. The temperature of water after 10 min is