# Water was poured into an aluminum pan weighing 300 g and a volume of 1 liter.

**Water was poured into an aluminum pan weighing 300 g and a volume of 1 liter. The initial temperature of the pot with water is 15 degrees. Heat the water in a saucepan to a boil. How much heat was required for this?**

Given:

m saucepan = 300 g = 0.3 kg;

V = 1 l = 0.001 m³;

T1 = 15 C = 288 K;

T2 = 100 C = 373 K;

from water = 4200 J / kg * C;

from aluminum = 920 J / kg * C;

Q =?

Solution.

The formula for heating a body has the form Q = cm * (T2-T1), but under the conditions of this problem, 2 objects are considered: a saucepan and water, which have different mass, specific heat and density. Hence, it is necessary to separately find the heating Q for both objects and then add them.

Q water = from water * m water * (T2 – T1);

Find m of water through the density formula:

m water = p water * V = 1000 * 0.001 = 1 kg.

Q water = 4200 J / kg * C * 1 kg * (373 K – 288 K) = 357 000 J.

Q castr = c alum * m castr * (T2 – T1) = 920 J / kg * C * 0.3 kg * (373 K – 288 K) = 23 460 J.

Q = Q water + Q castr = 357,000 J + 23 460 J ≈ 380.5 kJ.

Answer: ≈ 380.5 kJ.