Question

Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)

Answer 1

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

• Q=?
• m= 45 g
• c= 4.184 $\frac{J}{g*C}$
• ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J