Heat required by ice at 0°C to reach a temperature of 100°C = mL + mc∆θ 
= 1 × 80 + 1 × 1 × (100 – 0) = 180 cal
Heat available with 1 g steam to condense into 1 g of water at 100°C = 536 cal.
Obviously the whole steam will not be condensed and ice will attain a temperature of 100°C; so the temperature of mixture = 100 °C

Correct Option: C

Heat required by ice at 0°C to reach a temperature of 100°C = mL + mc∆θ 
= 1 × 80 + 1 × 1 × (100 – 0) = 180 cal
Heat available with 1 g steam to condense into 1 g of water at 100°C = 536 cal.
Obviously the whole steam will not be condensed and ice will attain a temperature of 100°C; so the temperature of mixture = 100 °C

Amount of energy radiated ∝ T⁴

Correct Option: D

Amount of energy radiated ∝ T⁴

Thermal capacity = ms = 40 × 0.2 = 8 cal/°C
= 4.2 × 8 J = 33.6 joules/°C

Correct Option: D

Thermal capacity = ms = 40 × 0.2 = 8 cal/°C
= 4.2 × 8 J = 33.6 joules/°C

Let the final temperature  be T
Heat gained by ice  = mL + m × s × (T – 0)
= 10 × 80 + 10 × 1 × T
Heat lost by water = 55 × 1× (40 – T)
By using law of calorimetery,
800 + 10 T = 55 × (40 – T)
⇒ T = 21.54°C = 22°C

Correct Option: B

Let the final temperature  be T
Heat gained by ice  = mL + m × s × (T – 0)
= 10 × 80 + 10 × 1 × T
Heat lost by water = 55 × 1× (40 – T)
By using law of calorimetery,
800 + 10 T = 55 × (40 – T)
⇒ T = 21.54°C = 22°C