When 100 gm of steam is added to 0.4 kg of water . The resultant temperature of the
mixture is
Answers
20.83 °C
Understanding :
1) Net Heat Change in this Process = 0
2) ice will be converted to water, then all will come at a equilibrium. Temperature.
Units are in SI system.
Steps:
1) Given,
Mass of Caloriemeter, m = 100g = 0.1kg
Specific Heat of Caloriemeter, S (C) = 0.1 units
Mass of liquid, m. = 0.25 kg
Specific Heat of Liquid, s(L) = 0.4 units
Mass of Ice, = 0.01kg
Latent heat of fusion of Ice, L(f) = 80 units
Specific heat of water, S(w) = 1 units.
2) Heat due to fusion = m L
Heat due to change in T = ms* del(T)
Let the final Equilibrium Temperature be T. Celsius.
Q(net ) = Q(1) + Q(2) + Q(3) + Q(4)
0 = 0.1 * 0.1 (T-30) + 0.25 *0.4 (T-30) + 0.01 * 80
+ 1* 0.01 (T-0)
Multiply by 100 on both sides,
=>0 = T-30 + 10(T-30) + 80 + T
=> 12 T = 250
=> T = 20.83 °C
Therefore, Final Temperature of mixture = 20.83°C
Follow me