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You weighed your empty insulated cup and found that the mass of the cup was 3.2 g. You add some hot water and find that the total mass increases to 28.1 g. You got some ice from the bus tub, and found that it was wet. After drying it carefully, you take the temperature of the hot water in the cup and find that its temperature is 60.7 °C and then add the ice. The temperature of the water dropped, and came to equilibrium at 35.1 °C. At the end of the experiment you re-weigh the cup and find that the mass has increased to 33.0 g.What was the initial temperature of the ice?What is your experimental value for the heat of fusion?

Respuesta :

batolisis

Answer:

Initial temperature of ice for this experiment = 0⁰c

heat of fusion = 95 cal

Explanation:

mass of empty cup = 3.2 g

mass of cup filled with hot water = 28.1 g

hence : mass of hot water = 28.1 - 3.2 = 24.9 g

mass of hot water + ice + empty cup = 33.0 g

hence : mass of ice = 33.0 - 24.9 - 3.2 = 4.9 g

temperature of hot water = 60.7⁰c

when ice was added ( temperature of mixture ) = 35.1⁰c

A) Initial temperature of the ice

initial temperature of ice is  0⁰c

B) experimental value for heat of fusion

we have the apply the law of energy conservation

= heat lost(hot water ) = heat gain by ice      s = specific heat of water = 1 cal

= M*s*ΔT = Mice * Hf + Mice *s *T

= 24.9 * 1 * ( 60.7 - 35.1 ) = 4.9 * Hf + 4.9 * 1 * 35.1

= 24.9 ( 25.6 ) = 4.9 hf + 171.99

Hf ( heat of fusion ) = 465.45 / 4.9 = 94.99 ≈ 95 cal

mickymike92

The Initial temperature of ice for this experiment is 0⁰c

The experimental heat of fusion of ice is 95.1 cal/g

Temperature of wet ice

Wet ice means ice with water on top of it or ice that is melting.

Since the ice was wet or melting, it means that its temperature was 0⁰c

Initial temperature of ice for this experiment = 0⁰c

Experimental value of the heat of fusion of ice.

  • Heat of fusion is the amount of heat energy required to melt a given amount of a solid at its melting point.

Using the law of conservation of heat

  • Heat lost by hot water = Heat gained by ice

initial temperature of ice is  0⁰c

mass of empty cup, m1 = 3.2 g

mass of cup filled with hot water, m2 = 28.1 g

mass of hot water, m3 = m2 - m1

mass of hot water = 28.1 - 3.2

mass of hot water = 24.9 g

mass of hot water + ice + empty cup, m4 = 33.0 g

mass of ice, m5 = m4 - m2

mass of ice = 33.0 - 28.1 = 4.9 g

temperature of hot water = 60.7⁰c

final temperature of ice + hot water mixture = 35.1⁰c

Initial temperature of the ice = y

Heat lost by hot water, H = mcΔT

where m = mass of hot water

c = specific heat of water = 1 cal

ΔT = temperature change

H = 24.9 × 1 × {35.1 - 60.7}

H = 637.44

Therefore, Heat gained by ice = 637.44

  • Heat gained by ice = Heat gained in changing from ice at 0°C to water at 0°C + Heat gained in changing from water at 0°C to water at 35.1°C

Heat gained in changing from ice at 0°C to water at 0°C = m×Hf

  • where m is mass of ice
  • Hf is heat of fusion of ice

Heat gained in changing from water at 0°C to water at 35.1°C = mcΔT

where m = mass of water at 0°C

c = specific heat of water = 1 cal

ΔT = temperature change

Heat gained by ice = 4.9 × Hf + 4.9 × 1 × {35 - 0}

Heat gained by ice = 4.9 × Hf + 171.5

637.44 - 171.5 = 4.9 × Hf

Hf = 465.94/4.9

Hf = 95.1 cal/g

Therefore, Heat of fusion of ice = 95.1 cal/g

Learn more about heat of fusion at: https://brainly.com/question/87248

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