# Ice weighing 5 kg at a temperature of -20 ° C was brought into the room, after which the ice melted.

Ice weighing 5 kg at a temperature of -20 ° C was brought into the room, after which the ice melted. The resulting water was heated to 26 ° C. How much heat did it take?

Given:
m = 5 kg
t1 = – 20C
t = 0 C – ice melting temperature
t2 = 26 C
Сl = 2100 J / kg * С – specific heat capacity of ice
Sv = 4200 J / kg * C – specific heat of water
L = 330,000 J / kg – specific heat of fusion
Q -?
Q1 = Сл * m * (t-t1) – the amount of heat for heating ice
Q2 = L * m – the amount of heat required for melting
Q3 = Sv * m * (t2-t) – the amount of heat required to heat water
Q = Q1 + Q2 + Q3
Q1 = 2100 * 5 * (0 + 20) = 210,000 J
Q2 = 330,000 * 5 = 1,650,000 J
Q3 = 4200 * 5 * (26-0) = 546000 J
Q = 546000 + 1650000 + 2100000 = 2406000 J = 2406 kJ One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.