When 300. cal of energy is lost from a 125 g object, the temperature decreases from 45.0°C to 40.0°C. What is the specific heat of this object in Cal/g ‘C?

The answer to your question is C = 0.48 cal/g°C

Explanation:

Data

Heat = Q = 300 cal

mass = m = 125 g

temperature 1 = T1 = 45°C

temperature 2 = T2 = 40°C

Specific heat = C = ? cal/g°C

Formula

    H = mC(T2 - T1)

-Solve for Specific heat (C)

    C = H / m(T2 - T1)

-Substitution

    C = 300 / 125(40 - 45)

-Simplification

    C = -300 / 125(-5)

    C = -300 / -625

-Result

    C = 0.48 cal/g°C

b i think it’s a water wheel turning in a stream

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

plz give brainliest