2 kg of ice at – 20°c is mixed with 5 kg of water at 20°c in an insulating vessel having a negligible heat capacity. calculate the final mass of water remaining in the container. it is given that the specific heats of water and ice are 1 kcal/kg per °c and 0.5 kcal/kg/°c while the latent heat of fusion of ice is 80 kcal/kg :-
Answers
The final mass of water in the container will be 6 kg
Explanation:
Heat lost by water when its temperature drops from 20°C to 0°C
= msΔt
= 5 x 1 x 20
= 100 kcal
Heat gain by water when its temperature rises from -20°C to 0°C
= msΔt
= 2 x 0.5 x 20
= 20 kcal
Hence the remaining heat will be used to melt the ice into water
remaining heat = 100 - 20 = 80 kcal
given the latent heat of fusion of ice is 80 kcal/kg
let m kg of ice is converted to water, hence
m x latent heat = heat remained
=> m x 80 = 80
=> m = 1 kg
Hence the final mass of water will be 5+1 = 6 kg
Answer:
Explanation:
. Given: Mass (m) = 5 kg
Temp. difference (∆T) = 100 - 20 = 80°C
Specific heat capacity (c) = 1 kcal/kg°C
To find: Heat energy (Q) = ?
Formula: Q = mc∆T
Solution:Q = 5 × 1 × 80
= 400 kcal