An unknown solid has a specific heat capacity of 0.727 J/gC. If a 12.3 grams sample is heated and increases in temperature from 22.4C to 62.1C, calculate how much heat was absorbed? You must show your work. I'm a little lost

355Joules

Explanation:

The amount of heat absorbed can be calculated using the formula:

Q = m × c × ∆T

Where;

Q = amount of heat absorbed in joules

m = mass of substance in grams

c = specific heat capacity of solid (0.727 J/g°C)

∆T = final temperature - initial temperature (°C)

According to the information provided in this question;

Q = ?

m = 12.3g

∆T = 62.1°C - 22.4°C = 39.7°C

c = 0.727 J/g°C

Using Q = m × c × ∆T

Q = 12.3 × 0.727 × 39.7

Q = 355.001

Q = 355Joules

Hence, 355J of heat was absorbed

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

points

explanation: