Question

How many kilojoules of heat energy are absorbed when 98.5 g of water are heated from 24.5 oC to 48.8 oC?

Answer 1

Answer:

10.01461 kilojoules of thermal energy are absorbed when 98.5 g of water is heated from 24.5 ° C to 48.8 ° C

Explanation:

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change) causing a change in temperature.

The equation that allows to calculate heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature.

In this case:

• c=4.184 $\frac{J}{g*C}$
• m= 98.5 g
• ΔT= Tfinal - Tinitial= 48.8°C - 24.5°C= 24.3 °C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ *98.5 g* 24.3 °C

Solving:

Q=10,014.61 J

Since 1 J is equal to 0.001 kJ, then you can apply the following rule of three: if 1 J is equal to 0.001 kJ, then 10,014.61 J is equal to how many kJ?

$kJ=\frac{10,014.71 J*0.001 kJ}{1 J}$

kJ=10.01461 kJ

10.01461 kilojoules of thermal energy are absorbed when 98.5 g of water is heated from 24.5 ° C to 48.8 ° C