A 2.50 g sample of zinc is heated, then placed in a calorimeter containing 65.0 g of water. Temperature of water increases from 20.00 oC to 22.50 oC. The specific heat of zinc is 0.390 J/goC. What was the initial temperature of the zinc metal sample? (final temperatures of zinc and water are the same)

719.83°C

Explanation:

The heat that the sample of Zinc gives is equal to the heat that water is absorbing. That is:

C(Zn) * m(Zn) * ΔT(Zn) = C(H2O) * m(H2O) * ΔT(H2O)

Where:

C is specific heat (Zn: 0.390J/g°C; H2O: 4.184J/g°C)

m is mass (Zn: 2.50g; H2O: 65.0g)

ΔT (Zn: ?; H2O: (22.5°C - 20.0°C = 2.50°C)

Replacing:

0.390J/g°C * 2.50g * ΔT(Zn) = 4.184J/g°C * 65.0g * 2.50

ΔT(Zn) = 697.33°C

As final temperature of Zn is 22.50°C, initial temperature is:

Initial temperature: 697.33°C + 22.50°C

719.83°C

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer