I heat 29.292 g of an unknown metal up to 99.9 °C. While it is heating, I weigh out 27.777 g of water, and find its initial
temperature is 22.1 "C. When I mix the metal and water in an insulated container, the temperature of the mixture rises
to 29.3 °C.
What is the most likely specific heat of the metal?
Select one:
a. 4.8 (104) cal / (g *C)
b. 9.7 (102) cal/(g*C)
c. 7.9 (10³) cal/ (g*C)
Od: 0.13 cal / (g°C)
e. 9.8 (105) cal / (g°C)

Answers

Answer 1

The specific heat capacity of the metal, given that 27.777 g of water at 22.1 °C was mixed with the metal is 9.7×10⁻² Cal/gºC

How do I determine the specific heat capacity of the metal?

Step 1: Obtain the heat absorbed by the water. This is shown below:

Mass of water (M) = 27.777 gInitial temperature (T₁) = 22.1 °CFinal temperature (T₂) = 29.3 °CTemperature change (ΔT) = 29.3 - 22.1 = 7.2 °CSpecific heat capacity of water (C) = 1 Cal/gºC Heat absorbed (Q) =?

Q = MCΔT

= 27.777 × 1 × 7.2

= 199.9944 Cal

Step 2: Determine the specific heat capacity of the metal using the heat absorbed by the water. Details below:

Heat absorbed by water (Q) = 199.9944 CalHeat released by metal (Q) = -199.9944 CalMass of metal (M) = 29.292 gInitial temperature (T₁) = 99.9 °CFinal temperature (T₂) = 29.3 °CTemperature change (ΔT) = 29.3 - 99.9 = -70.6 °CSpecific heat capacity of metal (C) = ?

Q = MCΔT

-199.9944 = 29.292 × C × -70.6

-199.9944 = -2068.0152 × C

Divide both sides by -2068.0152

C = -199.9944 / -2068.0152

= 9.7×10⁻² Cal/gºC

Thus, the specific heat capacity of the metal is 9.7×10⁻² Cal/gºC. None of the options are correct.

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