Answer:
Explanation:
Here, we want to get the amount of heat transferred
We can get that with the following formula:
[tex]Q\text{ = mc}\Delta\theta[/tex]
where:
[tex]\begin{gathered} Q\text{ = quantity of heat} \\ m\text{ = mass of the copper - 500g} \\ \Delta\theta\text{ = (25-35) = -10 deg. celsius} \\ c\text{ = 0.385 J/gC} \end{gathered}[/tex]
Substituting these values, we have it that:
[tex]Q\text{ = 500}\times0.385\times(-10\text{ )= -1,925 J}[/tex]
We have a negative answer as we are moving from a higher to a lower temperature which indicates a heat loss
