When 66.0 g of an unknown metal at 28.5 $\circ$ C is placed in 83.0 g H₂O at 78.5 $\circ$ C,the water temperature decreases to 75.9 $\circ$ C.What is the specific heat capacity of the metal? The specific heat capacity of water is 4.184 J/g.K.

Question

Quizplus · Accepted Answer

The heat lost by the metal is equal to the heat gained by the water:
(metal mass) x (metal specific heat capacity) x (change in temperature) = (water mass) x (water specific heat capacity) x (change in temperature)
(66.0 g) x (metal specific heat capacity) x (75.9°F - 28.5°F) = (83.0 g) x (4.184 J/g°C) x (78.5°F - 75.9°F)
Simplifying and solving for the metal specific heat capacity:
(metal specific heat capacity) = [(83.0 g) x (4.184 J/g°C) x (78.5°F - 75.9°F)] / [(66.0 g) x (75.9°F - 28.5°F)]
(metal specific heat capacity) = 0.29 J/g°C, which is equivalent to 0.29 J/g°F. Therefore, the correct answer is B.

When 660 G of an Unknown Metal at 28 ∘\circ∘ C Is Placed in 83

When 660 G of an Unknown Metal at 28 $\circ$ C Is Placed in 83