In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. she heats 19.6 grams of zinc to 98.37°c and then drops it into a cup containing 82.9 grams of water at 24.16°c. she measures the final temperature to be 25.70°c. assuming that all of the heat is transferred to the water, she calculates the specific heat of zinc to be __ j/g°c. g

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

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

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

the dissolution of polymer can be thermodynamically described in 2 steps.