# A 60 g piece of metal at an initial temperature of 92°C is transferred into 70 g of water initially at 24.5°C. The final temperature of the water and the metal is 40°C. How do you determine the specific heat of the metal?

## Assume perfect heat transfer.

Mar 16, 2017

$\frac{- 1.45362 J}{g r a m {C}^{o}}$

#### Explanation:

${q}_{\text{gain" = q_"lost}}$

(60g)(92^o - 40^oC)(x) = (70g)(24.5^oC - 40^oC)(4.18)"

4.18 is specific of water. Solve for x

$3120 x = - 4535.3$

$x = \frac{4535.3}{3120} \mathmr{and} \frac{45353}{31200}$

$x = - 1.45362$

This is the specific heat