Question

The specific heat of copper is $0.385$ J/(g. ${}^{o}C$). If $34.2$ g of copper, initially at ${25}^o$C, absorbs $4.689$ kJ, what will be the final temperature of the copper?

• A. ${25.4}^o$C
• B. ${27.8}^o$C
• C. ${356}^o$C
• D. ${381}^o$C

Answer 1

The correct option is D ${381}^o$C

$q=mc\Delta T$

Given that ,

$q=4689J,m=34.2,c=0.385J{g}^{-}{C}^{-}$

Now,

$4689=\left(34.2\right)\left(0.385\right)(x-25)$

$x={381}^{\circ }C$