Question

How much heat is required to warm 50.0 g of ice from -10.0oC to 0.00oC, melt the ice, warm the water from 0.00oC to 100.0oC, boil the water, and heat the steam to
120.0oC ?
a 209,000 J
b 199,000 J
c 1.67 x 106 J
d 152,000 J

Answer 1

The total heat required to convert the ice to steam is 155,000 J.

The given parameters:

• Mass of the ice, m = 50 g
• Initial temperature of the ice, t = -10 ⁰C
• Final temperature of the ice, T = 0⁰C, 100⁰C and 120⁰C
• Specific heat capacity of water = 4.184 J/g⁰C
• Heat of fusion of ice, = 333.55 J/g
• Heat of vaporization, = 2,230 J/g

The heat required to raise the temperature to 0⁰C is calculated as;

$Q = mc\Delta t\\\\Q_1 = 50 \times 4.184 \times (0 - (-10))\\\\Q_1 = 2092 \ J$

The  heat required to melt the ice is calculated as follows;

$Q_2 = mL_f\\\\Q_2 = 50 \times 333.55 \\\\Q_2 = 16,677.5 \ J$

The heat raise the temperature to 100⁰C is calculated as;

$Q_3 = 50 \times 4.184 \times (100 - 0)\\\\Q_3 = 20,920 \ J$

The  heat required to boil the water is calculated as follows;

$Q_4 = mL_v\\\\Q_4 = 50 \times 2230\\\\4_4 = 111,500 \ J$

The heat raise the temperature to 120⁰C is calculated as;

$Q_5 = 50 \times 4.184 \times (120 - 100)\\\\Q_5 = 4,184 \ J$

The total heat required is calculated as follows;

$Q_t = Q_1 + Q_2 + Q_3 + Q_4 + Q _5 \\\\Q_t = 2092 + 16,677.5 + 20,920 + 111,500 + 4,184\\\\Q_t = 155,373.5 \ J\\\\Q_t \approx 155,000 \ J$

Learn more about heat capacity here: brainly.com/question/16559442