An unknown substance weighing 42.0g was heated from 36.0°c to 65.0°c. in this process, the substance absorbed 5895 j of energy. what is the specific heat of the substance?

4.838 J / g °C

Explanation:

To do this, we need to use the correct expression or formula that involves heat, mass and temperature.

Q = m * Cp * ΔT (1)

Where:

Q: heat or energy absorbed or released.

m: mass of the substance or compound.

Cp: specific heat of the substance or compound

ΔT: change of temperature of the substance

We have the energy absorbed which is 5,895 J, the mass of 42 g (or 0.042 kg) and the temperature. First let's calculate ΔT:

ΔT = T2 - T1

ΔT = 65 - 36 = 29 °C

Now, with expression (1), let's solve for Cp:

Cp = Q / m * ΔT

Replacing the data:

Cp = 5,895 / 42 * 29

Cp = 4.839 J / g °C

This is the specific heat of the substance

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)