A coffee-cup calorimeter contains 10.0 g of water at 59.00°C.If 3.00 g gold at 15.20°C is placed in the calorimeter,what is the final temperature of the water in the calorimeter? The specific heat of water is 4.18 J/g⋅°C;the specific heat of gold is 0.128 J/g⋅°C.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat released by the gold and gained by the water:
$q_{Au} = m_{Au}c_{Au}\Delta T_{Au} = 3.00g 	imes 0.128 (J/g\cdot°C) 	imes (59.00°C-T_{final})$
$q_{H_2O} = m_{H_2O}c_{H_2O}\Delta T_{H_2O} = 10.0g 	imes 4.18 (J/g\cdot°C) 	imes (T_{final}-59.00°C)$

Since $q_{Au}=-q_{H_2O}$ (heat lost by one substance is equal to the heat gained by another), we can set them equal to each other and solve for Tfinal:

$3.00g 	imes 0.128 (J/g\cdot°C) 	imes (59.00°C-T_{final}) = 10.0g 	imes 4.18 (J/g\cdot°C) 	imes (T_{final}-59.00°C)$

Solving for Tfinal gives: 
$T_{final} = 58.60°C$

Therefore, the answer is B.

A Coffee-Cup Calorimeter Contains 10