An unknown amount of water was heated with 3.5 kJ, raising
its temperature from 26°C to
66°C. What was the mass of
the water?

20.9grams

Explanation:

Using the formula:

Q = m × c × ∆T

Where;

Q = amount of heat (joules)

m = mass of substance (g)

c = specific heat of water (4.184 J/g°C)

∆T = change in temperature (°C)

According to the provided information;

Q = 3.5kJ = 3.5 × 1000 = 3500J

m = ?

c of water = 4.184 J/g°C

∆T = 66°C - 26°C = 40°C

Using Q = m × c × ∆T

m = Q ÷ (c × ∆T)

m = 3500 ÷ (4.184 × 40)

m = 3500 ÷ 167.36

m = 20.9

mass = 20.9grams

the answers are

c and c