A 19.29 g of unknown metal is heated to 99.7 oC in a hot water bath. The metal s

Question

A 19.29 g of unknown metal is heated to 99.7 oC in a hot water bath. The metal sample is quickly transferred to a calorimeter containing 50.0 mL of water at 23.4 oC. The final maximum temperature for the mixture, from the graph, is 26.4 oC. What is the specific heat of the metal? (Density of water = 1.0 g/mL)
A 19.29 g of unknown metal is heated to 99.7 oC in a hot water bath. The metal sample is quickly transferred to a calorimeter containing 50.0 mL of water at 23.4 oC. The final maximum temperature for the mixture, from the graph, is 26.4 oC. What is the specific heat of the metal? (Density of water = 1.0 g/mL)
A 19.29 g of unknown metal is heated to 99.7 oC in a hot water bath. The metal sample is quickly transferred to a calorimeter containing 50.0 mL of water at 23.4 oC. The final maximum temperature for the mixture, from the graph, is 26.4 oC. What is the specific heat of the metal? (Density of water = 1.0 g/mL)

Explanation / Answer

Q = mcT

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kgK), is a symbol meaning "the change in"

T = change in temperature (Kelvins, K)

Heat lost by metal = heat gained by water

Mass of Metal = 19.29 gm

Mass of water = 50 ml = 50 gm

Specific heat of water = 4.184 J g-1 Celcius-1

19.29 x Sp Ht x (99.7 - 26.4) = 50 x 4.184 x (26.4-23.4)

Specific Heat of metal = 0.4438 J gram-1 Celcius-1

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