How much heat is required to warm 1.70 L of water from 26.0 ∘C to 100.0 ∘C ? (Assume a density of 1.0g mL ^-1 for the water.)

Question

Express your answer in joules to two significant figures.

Answer 1

The equation to use is q = mC∆T

q = heat = ?

m = mass = 1.70 L x 1000 ml / L x 1.0 g / ml = 1700 g

C = specific heat of water = 4. 184 J/gº

∆T = change in temperature = 100º - 26º = 74º

Solving for q, we have ...

q = (1700 g)(4.184 J/gº)(74º)

q = 526,347 J = 530,000 J (2 significant figures)