The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will
cause the ice to
A) change to water at a higher temperature
B) increase in temperature
C) change to water at the same temperature
D) decrease in temperature

Question

27-11-2019
Chemistry

contestada

The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will
cause the ice to
A) change to water at a higher temperature
B) increase in temperature
C) change to water at the same temperature
D) decrease in temperature

Respuesta :

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serverknock serverknock · Answer 1 · 2021-02-16T17:23:58+01:00

serverknock serverknock

16-02-2021

Answer:

D is the answer

Explanation:

Dont know

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boffeemadrid boffeemadrid · Answer 2 · 2021-11-10T10:00:14+01:00

The effect of heat added to ice is required.

The correct option is C) change to water at the same temperature

The heat of fusion of ice is 334 J/g. This means the heat required to convert 1 gram of ice to water is 334 J.

The one gram of ice at STP can only absorb 334 J before it starts melting.

The temperature of the water will remain constant.

The melting point and freezing point of water is the same [tex](0^{\circ}\text{C})[/tex]

The addition of energy to the solid phase of water (ice) causes the particles in it to have higher potential energy causing it to melt.

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The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will cause the ice to A) change to water at a higher temperature B) increase in temperature C) change to water at the same temperature D) decrease in temperature

Respuesta :

Otras preguntas

The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will
cause the ice to
A) change to water at a higher temperature
B) increase in temperature
C) change to water at the same temperature
D) decrease in temperature