
Water at $0^{\circ} C$ cools a soft drink bottle....... than Ice at $0^{\circ} C$.
Answer
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Hint: Ice in a cool drink is more beneficial at cooling it down than water at $0^{\circ} C$, which follows. Water that is cold can only receive heat energy to become warmer. Ice consumes heat energy, changing to water at $0^{\circ} C$ and then receiving further heat energy. So, the ice water receives more heat energy and is more productive in keeping things cool.
Complete step-by-step solution:
Because the $0^{\circ} C$ water holds more energy than $0^{\circ} C$ ice, it needs to change ice into water. Specifically, it needs $334$ joules per gram of ice. This implies that if we insert ice in our soft drink, $334$ joules of heat per gram of ice must be transported from the soft drink to the ice to soften it. That waste of energy will gain the soft drink cooler. Since some joules are needed to cool $1$ gram of water $1^{\circ} C$, we can observe that by melting just $1$ gram ice, about $80$ grams soft drink will be chilled by $1^{\circ} C$.
Contrast that to $0^{\circ} C$ water. Set that in a soft drink, and each gram of cold water will prepare $1^{\circ} C$ more heated for each gram of soft drink that makes $1^{\circ} C$ more frozen.
Because the distinction between a liquid at its transformation temperature and a solid at the equivalent is the Latent heat, this is the origin of refrigeration and AC. When you burn a pot of water to $100$ degrees Celsius, it begins its transformation temperature, where the water molecules split free of the bonds that keep them as a liquid. During this period, the temperature keeps at precisely $100$ degrees Celsius until the water has transformed from a liquid to a gas. Meanwhile, the quantity of energy has been rising to attain this state, despite the equal temperature.
A similar is true when we are speaking of the distinction between solid ice and water; there is a transformation phase wherever the temperature is equal while the ice is melting, but there is a quantity of energy needed to obtain this transition, so if we need cool drinks, begin with ice.
Water at $0^{\circ} C$ cools a soft drink bottle less than Ice at $0^{\circ} C$.
Note: For ice to melt, it should absorb heat. In order for water to cool, it has to deliver heat. If the mixture is neither receiving or discharging heat, ice can't vanish and water can't chill, so we have a constant and unchanging system. Interestingly, if we give heat, the melting ice will receive the attached heat, holding the mixture. If we extract heat, the freezing water will deliver heat, maintaining the temperature of the mixture fixed. As great as there's both water and ice combined together, the temperature doesn't vary.
Complete step-by-step solution:
Because the $0^{\circ} C$ water holds more energy than $0^{\circ} C$ ice, it needs to change ice into water. Specifically, it needs $334$ joules per gram of ice. This implies that if we insert ice in our soft drink, $334$ joules of heat per gram of ice must be transported from the soft drink to the ice to soften it. That waste of energy will gain the soft drink cooler. Since some joules are needed to cool $1$ gram of water $1^{\circ} C$, we can observe that by melting just $1$ gram ice, about $80$ grams soft drink will be chilled by $1^{\circ} C$.
Contrast that to $0^{\circ} C$ water. Set that in a soft drink, and each gram of cold water will prepare $1^{\circ} C$ more heated for each gram of soft drink that makes $1^{\circ} C$ more frozen.
Because the distinction between a liquid at its transformation temperature and a solid at the equivalent is the Latent heat, this is the origin of refrigeration and AC. When you burn a pot of water to $100$ degrees Celsius, it begins its transformation temperature, where the water molecules split free of the bonds that keep them as a liquid. During this period, the temperature keeps at precisely $100$ degrees Celsius until the water has transformed from a liquid to a gas. Meanwhile, the quantity of energy has been rising to attain this state, despite the equal temperature.
A similar is true when we are speaking of the distinction between solid ice and water; there is a transformation phase wherever the temperature is equal while the ice is melting, but there is a quantity of energy needed to obtain this transition, so if we need cool drinks, begin with ice.
Water at $0^{\circ} C$ cools a soft drink bottle less than Ice at $0^{\circ} C$.
Note: For ice to melt, it should absorb heat. In order for water to cool, it has to deliver heat. If the mixture is neither receiving or discharging heat, ice can't vanish and water can't chill, so we have a constant and unchanging system. Interestingly, if we give heat, the melting ice will receive the attached heat, holding the mixture. If we extract heat, the freezing water will deliver heat, maintaining the temperature of the mixture fixed. As great as there's both water and ice combined together, the temperature doesn't vary.
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