Qstnid -Iv-4-four 2 Cm &Times; 2 Cm &Times; 2 Cm Cubes Of Ice Are Taken Out From A Refrigerator And Are Put In 200 Ml Of A Drink At 10°c. (A) Find The Temperature Of The Drink When Thermal Equilibrium Is Attained In It. (B) If The Ice Cubes Do Not Melt Completely, Find The Amount Melted. Assume That No Heat Is Lost To The Outside Of The Drink And That The Container Has Negligible Heat Capacity. Density Of Ice = 900 Kg M-3, Density Of The Drink = 1000 Kg M-3, Specific Heat Capacity Of The Drink = 4200 J Kg-1 K-1, Latent Heat Of Fusion Of Ice = 3.4 &Times; 105 J Kg-1. (A) Find The Temperature Of The Drink When Thermal Equilibrium Is Attained In It. (B) If The Ice Cubes Do Not Melt Completely, Find The Amount Melted. Assume That No Heat Is Lost To The Outside Of The Drink And That The Container Has Negligible Heat Capacity. Density Of Ice = 900 Kg M-3, Density Of The Drink = 1000 Kg M-3, Specific Heat Capacity Of The Drink = 4200 J Kg-1 K-1, Latent Heat Of Fusion Of Ice = 3.4 &Times; 105 J Kg-1. - 25: Calorimetry

Home Learn Fast select Course

Show Unit/Chapter Change Unit Change Subject Last Menu

NEET-XII-Physics

25: Calorimetry

with Solutions - page 3

Qstn# iv-4 Prvs-Qstn Next-Qstn

#4
Four 2 cm × 2 cm × 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m^-3, density of the drink = 1000 kg m^-3, specific heat capacity of the drink = 4200 J kg^-1 K^-1, latent heat of fusion of ice = 3.4 × 10⁵ J kg^-1. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m^-3, density of the drink = 1000 kg m^-3, specific heat capacity of the drink = 4200 J kg^-1 K^-1, latent heat of fusion of ice = 3.4 × 10⁵ J kg^-1.
Ans : (a) Given:
Number of ice cubes = 4
Volume of each ice cube = (2 × 2 × 2) = 8 cm³
Density of ice = 900 kg m^-3
Total mass of ice, m_i = (4×8 ×10^-6 ×900) = 288×10^-4 kg
Latent heat of fusion of ice, L_i = 3.4 × 10⁵ J kg^-1
Density of the drink = 1000 kg m^-3
Volume of the drink = 200 ml
Mass of the drink = (200×10^-6)×1000 kg
Let us first check the heat released when temperature of 200 ml changes from 10^oC to 0^oC.
H_w = (200×10^-6)×1000×4200×(10-0) = 8400 J
Heat required to change four 8 cm³ ice cubes into water (H_i) = m_iL_i = (288×10^-4)×(3.4×10⁵) = 9792 J
Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( H_i > H_w ), some ice will remain solid and there will be equilibrium between
ice and water. Thus, the thermal equilibrium will be attained at 0^o C. (b) Equilibrium temperature of the cube and the drink = 0°C
Let M be the mass of melted ice.
Heat released when temperature of 200 ml changes from 10^oC to 0^oC is given by
H_w = (200×10^-6)×1000×4200×(10-0) = 8400 J
Thus,
M×(3.4×10⁵) = 8400 J
Therefore,
M = 0.0247 Kg = 25 g
Page No 47: (a) Given:
Number of ice cubes = 4
Volume of each ice cube = (2 × 2 × 2) = 8 cm³
Density of ice = 900 kg m^-3
Total mass of ice, m_i = (4×8 ×10^-6 ×900) = 288×10^-4 kg
Latent heat of fusion of ice, L_i = 3.4 × 10⁵ J kg^-1
Density of the drink = 1000 kg m^-3
Volume of the drink = 200 ml
Mass of the drink = (200×10^-6)×1000 kg
Let us first check the heat released when temperature of 200 ml changes from 10^oC to 0^oC.
H_w = (200×10^-6)×1000×4200×(10-0) = 8400 J
Heat required to change four 8 cm³ ice cubes into water (H_i) = m_iL_i = (288×10^-4)×(3.4×10⁵) = 9792 J
Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( H_i > H_w ), some ice will remain solid and there will be equilibrium between
ice and water. Thus, the thermal equilibrium will be attained at 0^o C. (b) Equilibrium temperature of the cube and the drink = 0°C
Let M be the mass of melted ice.
Heat released when temperature of 200 ml changes from 10^oC to 0^oC is given by
H_w = (200×10^-6)×1000×4200×(10-0) = 8400 J
Thus,
M×(3.4×10⁵) = 8400 J
Therefore,
M = 0.0247 Kg = 25 g
Page No 47:

#4-a
Find the temperature of the drink when thermal equilibrium is attained in it.
Ans : Given:
Number of ice cubes = 4
Volume of each ice cube = (2 × 2 × 2) = 8 cm³
Density of ice = 900 kg m^-3
Total mass of ice, m_i = (4×8 ×10^-6 ×900) = 288×10^-4 kg
Latent heat of fusion of ice, L_i = 3.4 × 10⁵ J kg^-1
Density of the drink = 1000 kg m^-3
Volume of the drink = 200 ml
Mass of the drink = (200×10^-6)×1000 kg
Let us first check the heat released when temperature of 200 ml changes from 10^oC to 0^oC.
H_w = (200×10^-6)×1000×4200×(10-0) = 8400 J
Heat required to change four 8 cm³ ice cubes into water (H_i) = m_iL_i = (288×10^-4)×(3.4×10⁵) = 9792 J
Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( H_i > H_w ), some ice will remain solid and there will be equilibrium between
ice and water. Thus, the thermal equilibrium will be attained at 0^o C.

#4-b
If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m^-3, density of the drink = 1000 kg m^-3, specific heat capacity of the drink = 4200 J kg^-1 K^-1, latent heat of fusion of ice = 3.4 × 10⁵ J kg^-1.
Ans : Equilibrium temperature of the cube and the drink = 0°C
Let M be the mass of melted ice.
Heat released when temperature of 200 ml changes from 10^oC to 0^oC is given by
H_w = (200×10^-6)×1000×4200×(10-0) = 8400 J
Thus,
M×(3.4×10⁵) = 8400 J
Therefore,
M = 0.0247 Kg = 25 g
Page No 47: