Question

Please with steps

Specific heat of ice = 0.48 cal/g · °C

Specific heat of liquid water = 1.00 cal/ g · °C

Specific heat of steam = 0.48 cal/g · °C

Heat of vaporization = 540 cal/g

Heat of fusion of ice = 80. cal/g

How much total heat in calories is required to raise the temperature of 2.30 g of ice at −19.0°C to water vapor at 121°C?

Total heat = ................cal

Answer 1

q₁ = heat required to warm the ice to 0.00 °C.

$q_1 = mcΔT = 2.3 \times 0.48 \times 19 = 20.976 \;cal$

q₂ = heat required to melt the ice to water at 0.00 °C.

$q_2 = mΔH_{fus} = 2.3 \times 80 = 184 \;cal$

q₃ = heat required to warm the water from 0.00 °C to 100.00 °C.

$q_3 = mcΔT = 2.3 \times 1.0 \times 100 = 230 \;cal$

q₄ = heat required to vapourize the water to vapour at 100 °C.

$q_4 = mΔH_{vap} = 2.3 \times 540 = 1242 \;cal$

q₅ = heat required to warm the vapour to 121 °C.

$q_5 = mcΔT = 2.3 \times 0.48 \times 21 = 23.184 \;cal$

$Total \;heat = q_1 + q_2 + q_3 +q_4 + q_5 \\ = 20.976 + 184 + 230 +1242 + 23.184 \\ = 1700.16 \;cal$

Answer: 1700 cal