Question

A 53 mL sample of water at an initial temperature of 41°C cooled down to a final temperature of 9°C. What is the amount of heat energy lost by the water? The specific heat of water is 4.184 J/g°C.

Answer 1

`Q\text{ = -7.2 KJ}`

Step-by-step explanation:

Here, we want to calculate the heat energy lost by the water

Mathematically, we can get that using the following mathematical relation:

`Q\text{ = mc}\Delta\theta`

where Q is the amount of heat

C is the specific heat capacity of water

delta theta is the temperature change which is the difference between the final and initial temperature

The mass of 53 mL of water is 0.0534 kg which is 53.4 g

Substituting the values, we have it that:

`\begin{gathered} Q\text{ = 53.4 }*\text{ 4.184}*(41-9) \\ Q\text{ = 7,149.6192 J = 7.2KJ} \end{gathered}`

Since it was cooled down, heat is lost which indicates a negative value