In an aluminum pan weighing 200 grams, 1.2 kg of water was heated from 20’C to 70’C. How much warmth went into this?

Given:
m1 = 200 grams (translated into SI) = 0.2 kg;
m2 = 1.2 kg;
t1 = 20’C;
t2 = 70’C;
c1 = 920 J / (kg * ‘C) – specific heat capacity of aluminum;
c2 = 4200 J / (kg * ‘C) – specific heat capacity of water.
To find:
Q-?
Decision:
To determine how much heat was required, we use the formula: Q = c * m * (t2 – t1), where c is the specific heat capacity, m is the body weight, t1 is the initial temperature, t2 is the final temperature.
Q = Q1 + Q2;
Q1 = c1 * m1 * (t2 – t1) = 920 * 0.2 * (70 – 20) = 9200 J – went to heat the pan;
Q2 = c2 * m2 * (t2 – t1) = 4200 * 1.2 * (70 – 20) = 252000 J – went to heating water;
Q = 9200 + 252000 = 261200 J;
Answer: 261200 J.