1 g of ice is mixed with 1 g of steam. At thermal equilibrium, the temperature of the mixture is

Question

1 g of ice is mixed with 1 g of steam. At thermal equilibrium, the temperature of the mixture is (A) 0°C (B) 100°C (C) 50°C (D) 55°C

Tardigrade Admin · Accepted Answer

Key concept According to principle of calorimetry states that total heat given by a hotter body is equal to the total heat received by colder body
i.e. heat lost by hotter body = heat gained by colder body
Heat required to melt

1 g

of ice at

0^{\circ} C

into

1 g

of water at

0^{\circ} C

is

80 c a l

.
Heat required to convert

1 g

of water at

0^{\circ} C

into

1 g

of water at

10 0^{\circ} C = 1 \times 1 \times 100 = 100 c a l

Heat required to condense

1 g

of steam

= 1 \times 540 c a l = 540 c a l

Clearly, whole of steam is not condensed. So, temperature of the mixture is

10 0^{\circ} C

.

Q. $1$ g of ice is mixed with $1 g$ of steam. At thermal equilibrium, the temperature of the mixture is

$0^{\circ} C$

$10 0^{\circ} C$

$5 0^{\circ} C$

$5 5^{\circ} C$

Q. 1 g of ice is mixed with 1g of steam. At thermal equilibrium, the temperature of the mixture is

0∘C

100∘C

50∘C

55∘C

Q. $1$ g of ice is mixed with $1 g$ of steam. At thermal equilibrium, the temperature of the mixture is

$0^{\circ} C$

$10 0^{\circ} C$

$5 0^{\circ} C$

$5 5^{\circ} C$