Question

Heat is added to 0.91 kg of ice at -6 °C. How many kilocalories are required to change the ice to steam at 138 °C? Hint: This involves two phase changes and heating the ice water and steam. Be sure to use the right latent heats and specific heats

Answer 1

given

mass = 0.91 kg

ice at - 6° C

ice to steam at 138 °C

using equation

Q = m C_ice dT + m L_f + m C_water dT + m L_v + m C_steam dT

= ( 0.91 x 2.108 x 6 ) + ( 0.91 x 334 ) + ( 0.91 x 4.184 x 100 ) + ( 0.91 x 2260 ) + ( 0.91 x 1.996 x ( 138 - 100 ) )

= ( 0.91 x 2.108 x 6 ) + ( 0.91 x 334 ) + ( 0.91 x 4.184 x 100 ) + ( 0.91 x 2260 ) + ( 0.91 x 1.996 x 38 )

= 0.91 x (( 2.108 x 6 ) + ( 334 ) + ( 4.184 x 100 ) + ( 2260 ) + ( 1.996 x 38 ))

= 0.91 x ( 12.648 + 334 + 418.4 +2260 + 75.848 )

= 0.91 x 3100.896

Q = 2821.815 k J

1 k J = 0.239006 k Cal

Q = 674.43 k Cal are required to change the ice to steam at 138 °C