Question #13200
25.0 g of an unknown metal at 90.0C is added to 50.0 g of water at 25.0 oC causing the temperature of the water rises to 29.8oC. What is the specific heat capacity and the identity of the metal?
Expert's answer
qmetal = qwater
By substitution, we have (metal values on the left, water values on the right):
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
(25.0)(90-29.8)(Cp) = (50.0)(29.8-25.0)(4.184 J g-1 °C-1)
Cp = 0.667 J g-1 °C-1
There is no metal with such Cp in tables.
By substitution, we have (metal values on the left, water values on the right):
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
(25.0)(90-29.8)(Cp) = (50.0)(29.8-25.0)(4.184 J g-1 °C-1)
Cp = 0.667 J g-1 °C-1
There is no metal with such Cp in tables.
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#339914 on Dec 2023
#339914 on Dec 2023