5.Sol:
[OFFLINE QUESTIONS]
1.Sol:
2.Sol:
3.Sol:4.Sol:5.Sol: The heat released by steam when it converts
from 100 C to 0 .C
Q 10 540 10 1 100
The mass of the ice that melts is
Q 6400 m 80 m 80 g
Water in calorimeter 500 80 10 590g
[OFFLINE QUESTIONS]
1.Sol: According to principle of calorimetry,
Heat lost = Heat gain
100 0.1 T 75 100 0.1 45 170 1 45
(^10) T (^75) 450 7650
(^) T 885 C
2.Sol: Assume that radius is decreased by R
mL AT
where m is the decrease in mass of the drop.
A is decrease in surface area. T is the
coefficient of surface tension.
4 R R L^24 T R^2 R R^2
R RL T R^2 ^2 R^22 R R R^2
R RL T R R^22 R is very small
R^2 T
L
3.Sol: U Q mc T
=100 10 ^3 4184 50 30 8.4kJ
4.Sol: Required work = Energy released
Here, Qmc dT
4 3
3
200.1 32 0.002
400 T dT kJ
Therefore, required work = 0.002 kJ.