In the laboratory a student determines the specific heat of a metal. he heats 19.5 grams of copper to 98.27 °c and then drops it into an insulated cup containing 76.3 grams of water at 24.05 °c. when thermal equilibrium is reached, he measures the final temperature to be 25.69 °c. assuming that all of the heat is transferred to the water, he calculates the specific heat of copper to j/g°c.

The specific heat of copper is 0.37 J/g°C

Explanation:

Step 1: Data given

Mass of copper = 19.5 grams

Initial temperature of copper = 98.27 °C

Mass of water = 76.3 grams

Initial temperature of water = 24.05 °C

Final temperature of water and copper = 25.69 °C

Step 2: Calculate specific heat of copper

Qgained = -Qlost

Q = m*c*ΔT

Qwater = -Qcopper

m(water) * c(water) * ΔT(water) = - m(copper) * c(copper) *ΔT(copper)

⇒ with m(water) = 76.3 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2-T1 = 25.69 - 24.05 = 1.64

⇒ with m(copper) = 19.5 grams

⇒ with c(copper) = TO BE DETERMINED

⇒ with ΔT(copper) = T2-T1 = 25.69 - 98.27 = -72.58

76.3 * 4.184 * 1.64 = - 19.5 * c(copper) * -72.58

523.552 = 1415.31 * c(copper)

c(copper) = 0.37 J/g°C

The specific heat of copper is 0.37 J/g°C

the correct answer is answer a. a mixture

no other options make sense

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  - sprinkles