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75.0 g sample of a metal at 65°C is added to 100.0 g of water at 20.0°C. The temperature of the water rises to 22.3 °C. Calculate the specific heat of the metal, assuming that all the heat lost by the metal is gained by the water.

Sagot :

The specific heat of the metal is 0.30J/g°C.

What is specific heat capacity?

The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).

Step 1: Data given

Mass of metal = 75.0 g

Temperature of the metal = 65°C

Mass of water = 100.0 g

Temperature of the water = 20.0°C

Final temperature of the water = 22.3 °C

The specific heat of water is 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Qlost = -Qgained

Q= m xCxΔT

Qmetal = -Qwater

m(metal) x c(metal) x ΔT(metal) = - m(water) x c(water) x ΔT(water

⇒m(metal) = the mass of metal = 75.0 g

⇒c(metal) = the specific heat of the metal = ?

⇒ΔT = The change of temperature = 22.3 - 65°C = -42.7 °C

⇒m(water) = the mass of water = 100.0 g grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT(water) = T2 - T1 = 22.3 - 20.0 = 2.3 °C

75.0 g x c(metal) x -42.7 °C= -100.0 g x  4.186 x 2.3

75.0 g x c(metal) x -42.7 °C = −962.78

c(metal) = 0.3006338798J/g°C =0.30J/g°C

The specific heat of the metal is 0.30J/g°C.

Learn more about specific heat capacity here:

https://brainly.com/question/1747943

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