The specific heat of water is 4.18 j g–1 k–1 and that of stainless steel is 0.51 j g–1 k–1. calculate the heat that must be supplied to a 750.0 g stainless steel vessel containing 800.0 g of water to raise its temperature from 20.0°c to the normal boiling point of water.

The heat that must be supplied is 298.1 kJ

Explanation:

We must apply the calorimetry formula to solve this.

Q = m . C . ΔT

where Q is heat, m is the mass, C is the specific heat and ΔT the difference between Final temperature and Initial temperature.

The normal boiling point of water is 100° so ΔT = (100° - 20°)

Notice that units in specific heat are J/ g.K so temperature should be in K but in this case K and C are the same, because it is a difference.

T°K = T°C + 273

ΔT = (100° - 20°) = 80°

ΔT = (373° - 293°) = 80°

750g . 0.51 J/gK . 80K + 800g . 4.18 J/gK . 80K = 298120 J

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"