What is the amount of heat energy absorbed when 36 grams of ice at -20oC is melted to water at 0oC?

Q = 1461.6 J

Explanation:

Given data:

Mass of ice = 36 g

Initial temperature = -20°C

Final temperature = 0°C

Amount of heat absorbed = ?

Solution:

specific heat capacity of ice is 2.03 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 = T2 - T1

ΔT =  0°C - (-20°C)

ΔT = 20°C

Q = 36 g ×2.03 j/g.°C×20°C

Q = 1461.6 J

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g