Question

a calorimeter has mass 100 gram and specific heat 0.1 kilo calorie per kilogram degree Celsius if contains 250 kilogram of liquid at 30 degree Celsius having specific heat of 0.4 kilo calorie per kilogram degree Celsius if we drop a space a piece of Ice and mass 10 gram at zero degree celsius what will be the temperature of the mixture

Answer 1

Final Answer :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

Answer 2

20.8°c is the answer of this question