Question

a 2.5 G sample of zinc is heated then placed in a calorimeter containing 65 G of water temperature of water increases from 20 degree Celsius to 22.5 degr

Answer 1

Explanation:

Data given: mass of zinc sample = 2.5 G

mass of water = 65 G

initial temperature of water = 20 degrees

final temperature of water = 22.5 degrees

ΔT = change in temperature of water is 2.50 degrees

specific heat capacity of zinc cp= 0.390 J/g°C

initial temperature of zinc sample = ?

cp of water = 4.186 J/g°C

heat absorbed = heat released (no heat loss)

formula used is,

q = mcΔT

q water = 65 x 4.286 x 2.5

q water = 696.15 J ---------->(1)

q zinc = 2.5 x 0.390 x (22.50- Ti) ---------->(2)

equating the two equations

696.15 = - 22.50+ Ti

Ti = 718.65 degrees is the initial temperature of zinc.