Alaboratory technician drops a 0.0850 kg sample of unknown material, at a temperature of 100.0∘c, into a calorimeter. the calorimeter can, initially at 19.0∘c, is made of 0.150 kg of copper and contains 0.200 kg of water. the final temperature of the calorimeter can is 26.1∘c.

1013.32 J/kg.K

Explanation:

The heat transferred by a changing in temperature without phase change can be calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat, and ΔT is the change in temperature (final - initial).

The values of c for water and copper can e found in thermodynamics tables:

cwater = 4.19x10³ J/kg.K

ccopper = 0.39x10³ J/kg.k

By the conservation of energy:

Qwater + Qcopper + Qmaterial = 0

0.200*4.19x10³*(26.1 - 19.0) + 0.150*0.39x10³*(26.1 - 19.0) + 0.085*c*(26.1 - 100) = 0

5949.8 + 415.35 - 6.2815c = 0

6.2815c = 6365.15

c = 1013.32 J/kg.K

a electronegativity decreases beacuse the size of nucleus is bigger.