Ice was placed in a thermos with water at a temperature of -10. The mass of water is 400 g
Ice was placed in a thermos with water at a temperature of -10. The mass of water is 400 g, the mass of ice is 100 g, the initial water temperature is 18 ° C. Determine the final temperature established in the thermos
Heating and melting ice requires heat Q1:
Q1 = cлm1 (tpl – tl) + Lm1,
where w is the specific heat capacity of ice,
m1 – mass,
tl – initial temperature,
tmelt – melting point,
L is the specific heat of fusion.
Q1 = 2050 J / (kg deg) * 0.1 kg * (0 ° C – (-10 ° C)) + 330 000 J / kg * 0.1 kg = 2050 J +33 000 J = 35 050 J …
When water is cooled to 0 ° C, it can give off heat Q2:
Q2 = cm2 (tv – 0 ° С) = 4200 J / (kg * deg) * 0.4 kg * 18 ° С = 30 240 J, where c is the specific heat capacity of water, tv is the initial temperature.
Answer: Temperature – 0 ° С, ice mass – 0.014 kg.
