Answer to Question #168034 in General Chemistry for Youmi

Question #168034

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


1
Expert's answer
2021-03-10T06:00:19-0500

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

"q_1 = mc\u0394T = 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 = m\u0394H_{fus} = 2.3 \\times 80 = 184 \\;cal"

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

"q_3 = mc\u0394T = 2.3 \\times 1.0 \\times 100 = 230 \\;cal"

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

"q_4 = m\u0394H_{vap} = 2.3 \\times 540 = 1242 \\;cal"

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

"q_5 = mc\u0394T = 2.3 \\times 0.48 \\times 21 = 23.184 \\;cal"

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

Answer: 1700 cal


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