Question

What is the maximum mass of water that can be heated from 308 K to 313 K by the addition of 1260 J. of heat?

Answer 1

Answer:

60 g

Explanation:

The formula relating mass (m), heat (q), and temperature change (ΔT) is

q = mCΔT     Divide both sides by CΔT and switch

m = q/(CΔT)

q = 1260 J

C = the specific heat capacity of water = 4.184 J·K⁻¹g⁻¹

ΔT = T₂ - T₁ = 313 K – 308 K = 5 K

m = 1260/(4.184 × 5)

m = 1260/21

m = 60 g