Question

When 66.0 g of an unknown metal at 28.5 °C is placed in a calorimeter filled with 83.0 g H20 at 78.5 °C, the water temperature decreases to 75.9 °C. Accounting for the heat capacity of the calorimeter, 33.5J/°C, calculate the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.184 J/g °C.

Answer 1

The specific heat capacity of the unknown metal is 0.288 J/g°C.

Step-by-step explanation:

To calculate the specific heat capacity of the metal, we can use the equation:
q = mcΔT
Where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the water:
m = 83.0 g, c = 4.184 J/g°C, and ΔT = 75.9°C - 78.5°C

= -2.6°C

qwater = (83.0 g)(4.184 J/g°C)(-2.6°C)

= -886.74 J
Since the water is losing heat, the metal is gaining the same amount of heat.

For the metal:
m = 66.0 g, ΔT = 75.9°C - 28.5°C = 47.4°C

qmetal = -qwater = 886.74 J
cmetal = qmetal / (mΔT) = 886.74 J / (66.0 g * 47.4°C)

= 0.288 J/g°C