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attashe74 [19]
3 years ago
8

a beaker with water and the surrounding air are all at degrees Celsius. After ice cubes are placed in the water, heat is transfe

rred from?
Chemistry
2 answers:
Yuki888 [10]3 years ago
8 0

(3) the water to the ice cubes
  This is so because the ice cubes are colder compared to their surrounding objects.
They absorb the heat from the water, cooling it, therefore vanishing(melting).

Another way to put it would be

The answer is (3) the water to ice cubes.

As the ice cubes should be at a temperature of about 0 degree ( freezing point) , at the same time the temperature of water is 24 degree. Thus, heat is transferred from water to ice cubes.

AfilCa [17]3 years ago
4 0

Answer:

(3) the water to the ice cubes.

Explanation:

There is some info missing. I believe this is the original question:

<em>A beaker with water and the surrounding air are all at 24°C. After ice cubes are placed in the water, heat is transferred from:(1) the ice cubes to the air(2) the beaker to the air(3) the water to the ice cubes(4) the water to the beaker.</em>

<em />

Heat is transferred from bodies at higher temperatures to bodies at lower temperatures.

<em>Heat is transferred from:</em>

<em>(1) the ice cubes to the air.</em> NO. Heat will be transferred from air (24°C) to the ice cubes (0°C).

<em>(2) the beaker to the air.</em> NO. The beaker and the air are at the same temperature.

<em>(3) the water to the ice cubes.</em> YES. Water (24°C) is at a higher temperature than ice (0°C).

<em>(4) the water to the beaker.</em> NO. The water and the beaker are at the same temperature.

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How much heat energy, in kilojoules, is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer
ANTONII [103]

Answer:

25.2 kJ of heat energy is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C

Explanation:

Data: mass of ice = m = 53.0 g

         Temperature of ice = T1 = -18.0 ∘C

         Temperature of water = T2 = 25.0 ∘C

          Change in Temperature = ΔT : T2-T1                                          

           Specific heat of ice = c(i) = 2.09 J/g . ∘C

          Specific heat of water = c(w) = 4.18 J/g . ∘C

          Enthalpy of fusion of water (when convert from ice to water) = ΔH(f) : 334 J/g

          Enthalpy of vapourization of water (when convert from water to gas) = ΔH(v) :2250 J/g

          Total heat required = q = ?

Solution:        Heat required to melt the ice

                      T1 = -18 ∘C

                      T2 = 0 ∘C

                      Q1 = m x c(i) x ΔT

                      Q1 = 53.0 x 2.09 x (0-(-18))

                      Q1 = 53.0 x 2.09 x 18

                      Q1 = 1993.86 J is required by ice to reach to its

                      melting point.

                      <u>Heat required to convert the ice into water</u>

                      Q2 = ΔH(f) x m

                      Q2 = 334 x 53.0 = 17702 J is required by ice to  

                      convert into water.

                      <u> Heat required by water to reach at 25 ∘C</u>

                       T1 = 0 ∘C

                       T2 = 25 ∘C

                       Q3 = m x c(w) x ΔT

                       Q3 = 53.0 x 4.18 x (25-0)

                       Q3 = 5538.5 J is required by water to reach at  

                       25∘C .

                       <u>Total heat required by 53.0 g of ice at −18.0 ∘C  </u>

                        <u>to water at </u>

                        <u>25.0 ∘C</u>

                        q = Q1 + Q2 + Q3

                        q = 1993.86 J + 17702 J + 5538.5 J

                        q = 25234.36 J is the total heat required 53.0 g  

                        of ice at −18.0 ∘C to water at 25.0 ∘C.

                        <u>Conversion from Joule (J) to Kilojoule (kJ)</u>

                          1000 J =  1 kJ

                          q = 25234.36 J/1000

                          q = 25.23436 kJ

                           <u>Conversion of heat in kJ to three significant   </u>

                           <u>figures</u>

                           q = 25.2 kJ

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