How much heat (in kj) is needed to convert 866 g of ice at −10°c to steam at 126°c? (the specific heats of ice and steam are 2.03 j/g · °c and 1.99 j/g · °c, respectively.)

2671.3 kJ

Explanation:

m = 866 g at - 10 degree C converts into steam at 126 degree C.

The following steps are there

(1) ice at - 10 degree C converts into ice at 0 degree C

H1 = m x specific heat of ice x ΔT

H1 = 866 x 2.03 x 10 = 17579.8 J

(ii) ice at 0 degree C converts into water at 0 degree C

H2 = m x Latent heat of fusion = 866 x 334 = 289244 J

(iii) water at 0 degree C converts into water at 100 degree C

H3 =  m x specific heat of water x ΔT

ΔH3 = 866 x 4.186 x 100 = 362507.6 J

(iv) water at 100 degree C converts into steam at 100 degree C

H4 = m x Lateant heat of vaporisation

H4 = 866 x 2260 = 1957160 J

(v) seam at 100 degree C converts into steam at 126 degree C

H5 = m x specific heat of steam x ΔT

H5 = 866 x 1.99 x 26 = 44806.84 J

The total heat required is

H = H1 + H2 + H3 + H4 + H5

H = 2671298.24 J = 2671.3 kJ

increase masses, decrease separation