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How much energy is required to raise the temperature of 74 g of water from 19 °C to 47 °C.

1 Answer

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Answer:

The energy required is 8660.96 J.

Step-by-step explanation:

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

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change). This heat is given by the expression:

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, you know:

  • C= 4.18
    (J)/(g*C)
  • m= 74 g
  • ΔT= Tfinal - Tinitial= 47 C - 19 C= 28 C

Replacing:

Q= 4.18
(J)/(g*C) *74 g* 28 C

Solving:

Q= 8660.96 J

The energy required is 8660.96 J.

User Priyanka Modi
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