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Alchen [17]
3 years ago
12

Ice wedging is brought about by ____.

Physics
1 answer:
grigory [225]3 years ago
7 0
<span>Ice wedging is caused by water freezing and thawing. During this process of conversion of the water into ice the water is expanded and crack in the ground is formed. These cracks are formed duing the winter, when the water freezes, expands and the enlarges the cracks. In summer, when the ice melts more water is accumulated in the cracks and the process repeats.</span>
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Question 17 A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off
ivanzaharov [21]

Complete Question:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

(Question attached)

Answer:

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C (rounded to 1 decimal place)

Explanation:

A calorimeter is used to measure the heat of chemical or physical reactions. The example given in the question is using the calorimeter to determine the specific heat capacity of iron.

When the system reaches equilibrium the iron and water will be the same temperature, T_{e}. The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.

Q=mc(T_{e}-T_{initial}) (Eq 1)

-Q_{iron}=Q_{water} (Eq 2)

Water:

m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C

Iron:

m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C

Substituting Eq 1 into Eq 2 and details extracted from the question:

-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})

-59.1*c_{iron}(27.6-85)=100.0*4.186(27.6-23)

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C

4 0
3 years ago
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