How much energy is required to change 110 grams of steam at 181 C to 40 C

Q = -64893.84 J

Explanation:

Given data:

Mass of steam = 110 g

Initial temperature = 181 °C

Final temperature = 40°C

Energy released = ?

Solution:

Energy will released in this process because temperature is going to be decreased.

Specific heat capacity of water is 4.184 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 40°C - 181 °C

ΔT = -141°C

Q =  110 g × 4.184 J/g.°C × -141°C

Q = -64893.84 J

28528289 29823282 8298232 39858

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.