# professional factory for recharged hot water bag Supply to Nicaragua

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Place of Origin:Zhejiang, China (Mainland) Brand Name:NUANYI Model Number:NY0077 classification:hand warmer stlye:with cover (can put hand in bag) Product name:electric hot water bag Item name:electric warm bag weight:1.4kg Style:custom size:26*18.5cm Material: Flannelette,PVC Voltage:220V(110v) Power:380w Charge time:8-15min Warm time:3-8 hours Manufactor: Cixi Jade Rabbit Electric Appliance Factory Approved Certificate:CE ROHS Wire length:75(80)cm Max...

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professional factory for recharged hot water bag Supply to Nicaragua Detail:

Place of Origin:Zhejiang, China (Mainland) |

Brand Name:NUANYI |

Model Number:NY0077 |

classification:hand warmer |

stlye:with cover (can put hand in bag) |

Product name:electric hot water bag |

Item name:electric warm bag |

weight:1.4kg |

Style:custom |

size:26*18.5cm |

Material: Flannelette,PVC |

Voltage:220V(110v) |

Power:380w |

Charge time:8-15min |

Warm time:3-8 hours |

Manufactor: Cixi Jade Rabbit Electric Appliance Factory |

Approved Certificate:CE ROHS |

Wire length:75(80)cm |

Maximumtemperature:75°c |

Reset temperature:35-40°c |

Pvc:double pvc |

Product detail pictures:

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Learn about energy diagrams, calculating molar heat of vaporization and fusion and molar heat of solution in this video!

Transcript

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In this diagram we have the temperature of water in degrees celsius on the side and energy added in kilojoules on the bottom. The graph starts by showing ice at a temperature below freezing. as we add heat the temperature goes up. Until we get to zero. At zero even though heat is being added, the temperature isn’t changing. This is because the heat is going into changing the phase of water from solid to liquid, not increasing the kinetic energy. The amount of energy required to either melt ice or freeze water is 6.01 kJ/mol. This is known as the delta H or enthalpy of fusion when it’s melting which is endothermic. It can also be called the enthalpy of solidification if we were freezing water into ice, which is an exothermic process. Once all the ice is melted, the heat will raise the temperature of the water until it hits 100 degrees celsius. This is the boiling point, or vaporization point of water. Now it will take 40.7 kJ/mol of water to turn the liquid into a vapor. This is the delta H of vaporization for water and it’s an endothermic process. If we were cooling steam into water, we would call it the enthalpy of condensation, which is exothermic, but it’s the same amount energy required as vaporization. The molar heat of vaporization is a much higher number than the molar heat of fusion for water. This means it takes more energy to change from a liquid to a gas than is does to change from a solid to a liquid in the case of water. After all the water becomes a vapor, the increased heat will continue to increase the temperature of the vapor. This is also why steam burns are worse than hot water burns, they have a much higher temperature! So how can you calculate all the energy required to go from ice to steam?

Calculate the amount of heat necessary to vaporize 125 g of ice at -10oC. We’ll need this information, q = mcat, the specific heat of water and ice and the enthalpy of vaporization and fusion for water. Now let’s break down the process.

First, we’ll calculate the energy needed to warm up the ice to 0oC. Since the temperature is changing as we add heat, we’ll use q=mcat. we’ll also need the specific heat capacity of ice, which is 2.1 j/goC. Now we plug in our data. so we have the mass of 125 grams, the specific heat and the final temperature of 0 minus the initial temperature of -10. Be careful with those double negatives, this creates a positive number. q = 2,625 J. The next step in the diagram is the melting of the ice.

Melting ice doesn’t change the temperature at all, so we can’t use q=mcat. Instead, we’ll use the enthalpy of fusion and dimensional analysis. 125g of water, times the molar mass of water times the enthalpy of fusion and we get 41.7 kJ or 41,700 J.

Heating water involves a change of temperature so we use q=mcat and the specific heat of liquid water. the plug in mass, specific heat and final temperature of 100 – initial temp of 0 and we get 52,300 Joules. Then we calculate the heat of vaporizing the water with the enthalpy of vaporization and dimensional analysis. 125 grams times the molar mass times enthalpy of vaporization and we get 283 kJ or 283,000 J. Finally you add together the values from each step. And we get 380,000 Joules or 380 kJ.

This process is really just like calorimetry, but instead of putting some object in water, you’ll mix two chemicals together in the calorimeter, or dissolve some chemical in water. You measure the temperature to it’s highest or lowest point to get the final temperature. Let’s try an example with the molar heat of solution. 0.333 mol of a solid was dissolved in 260 mL of water at 22.3 oC. After the solid had fully dissolved, the final temperature of the solution was 27.5 oC. What is the molar heat of solution of the substance? We start this problem just like the unknown metal calorimetry problem, we’ll calculate the q of water first because we have all the information.

Because 1 mL of water is equal to 1 gram, we can say that there are 260 grams of water, the specific heat is 4.184 j/goC, and the final – initial is 27.5 – 22.3. The heat is 5,700 Joules. The water absorbed heat which means the reaction gave off heat so we’ll make it a negative number.

We want to know the heat of dissolution in joules per mol so we just have to divide and we get negative 17000 j/mol . This chemical would make for a good hot pack because it’s exothermic when the attractions between the molecules or atoms are broken. Other chemicals are endothermic and would make good cold packs.