1. How much heat is needed to raise the temperature of 1 gallon of water to 37 t

Question

1. How much heat is needed to raise the temperature of 1 gallon of water to 37 to boiling? 2. Is it reasonable that this process would be done in one step or more than one step and why? 3. What is the apparent atomic mass of the metal and the identity of the metal? POST LAB QUESTIONS 1.-(5 pts How much heat is needed to raise the temperature of I gallon of water from 37 °C to boiling? Show your work. 2. (5 pts) A primitive way of heating water to boiling in a non-fireproof container was to heat stones or metal and then to place them in the container to boil the water. How large a piece of iron (OFe 0.447 J/g C) would be needed to boil 1.0 L of water if the iron is heated to 500 °C? Is it reasonable that this process would be done in one step or more than one step and why?0 3. (5 pts) A 53.72 g sample of an unknown metal was heated to 98.0 °C and then placed in a calorimeter with 60.0 g of water at 25.0 °C. if the final temperature in the calorimeter was 32.0 °C, what is the specific heat capacity of the metal in J/g C? What is the apparent atomic mass of the metal and the identity of the metal? 0 4. (5 pts) Instant ice packs get cold when an ampoule is broken inside the pack. contains water, describe what is happening and why it gets cold Assuming the ampoule

Explanation / Answer

Volume of water = 1 gallon = 3.785 L

mass of water = 3.785 Kg = 3785 grams

Initial temp = 37 0C

Final temperature = 100 0C

Change in temperature = 100 -37 = 63 0C

Heat required to raise the temperature = Mass of water X specific heat of water X change in temperature

Heat required = 3785 grams X 4.184 X 63 = 997695 Joules = 997.695 kJ

2) the temperature of iron = 500 C

Let us assume that the initial temperature of water = 25C

The heat loss by metal = heat gained by water

mass of metal X specific heat X change in temp = mass of water X specific heat X change in temp

MAss of iron X 0.447 x (500-100) = 1000 X 4.184 X (100-25)

Mass of iron = 1755 grams

3)

The heat loss by metal = heat gained by water

mass of metal X specific heat X change in temp = mass of water X specific heat X change in temp

53.72 X specific heat of metal X (98-32) = 60 X 4.184 X (32-25)

specific heat of metal = 0.496 J / g0C

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