# Find the specific heat of a substance weighing 200 g, which required 300 J

July 24, 2021 | education

| **Find the specific heat of a substance weighing 200 g, which required 300 J of heat to heat it from a temperature of 12˚C to a temperature of 16.4˚C.**

Given: m = 200 g = 0.2 kg; t1 = 12 ° C; t2 = 16.4 ° C; Q = 300 J.

Find: c.

Solution: the amount of heat is equal to the product of the mass of the substance by the change in temperature and by the specific heat of the substance. Q = c * m * ∆t; Q = c * (t2 – t1); c = Q / (m * (t2 – t1)).

c = 300 / (0.2 * (16.4 – 12)) = 300 / (0.2 * 4.4) = 300 / 0.88 ≈ 340.9 J / (kg * ° С).

The answer is 340, 9 J / (kg * ° С).

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.