Dantes Peak Essay Example

1. Make a copy of Table 1 for the assignment page. Data: |

Table 1 | Volume of cold water | 237 mL (1cup) | Volume of hot water | 237 mL (1 cup) | Beginning temperature of the cold water (oC) | 25oC | Beginning temperature of the hot water (oC) | 80oC | Final temperature of the cold/hot water mixture (oC) | 52oC | Change in temperature of the cold water (oC) | 27oC | Change in temperature of the hot water (oC) | 32oC | Mass of the cold water (kg) | .237kg | Mass of the hot water (kg) | . 237kg |

A. Use the beginning temperature of the hot and cold water and the final temperature of the mixture to calculate the change in temperature of the cold water and the change in temperature of the hot water. Record your values in Table 1. For example, the temperature of the cold water was raised from its beginning temperature to the final temperature of the mixture.

B. Since one milliliter (mL) of water has a mass of one gram (g), it is very easy to determine the mass of the cold and hot water. For example: If you have 100 mL of water, then the mass of the water is 100 g. You now know how many grams of cold and hot water you used in the experiment. Remembering that one kilogram of material is the same mass as 1000 g (1 kg = 1000 g), convert the mass of the hot and cold water to kilograms and record the values in Table

2. Use the equation Q = (m)(c)( T) to calculate the heat gained by the cold water. Show your work using the problem-solving method shown in previous rubrics. The specific heat for water (c) is 4186 J/(kg * Co).
Q=.237x4186x(27-25)
Q=1984.164J
3. Use the equation Q = (m)(c)( T) to calculate the heat "lost" by the hot water. Show your work using the problem-solving method shown in previous rubrics. The specific heat for water (c) is 4186 J/(kg * Co).
Q=.237X4186X(80-32)
Q=47619.936J

4. Compare the values for heat gain and heat loss in questions 2 and 3.
There was a lot more heat loss than gained.

5. In a closed