Question

At 1 atm, how much energy is required to heat 39.0 g of h2o(s at –24.0 °c to h2o(g at 121.0 °c?

Answer 1

q = mcΔt, q = energy [J] m = mass (of water) [g]; c = specific heat capacity of water [J g⁻¹ K⁻¹/°C⁻¹]; Δt = change in temperature [K/°C]
Δt = 121 - -24 = 145
q = 39 × 4.18 × 145
q = 23637.9 J