EXPERIMENT: 9 SPECIFIC HEAT OF ANTIFREEZE Materials Required: polystyrene cups a

Question

EXPERIMENT: 9 SPECIFIC HEAT OF ANTIFREEZE Materials Required: polystyrene cups anti-freeze (Prestone) Note: Ethylene glycol can be used in the place of anti-freeze. SAFETY PRECAUTIONS: EYE PROTECTION MUST BE WORN AT ALL TIMES IN LABORATORY. SPILLED ANTIFREEZE SHOULD BE WASHED IMMEDIATELY FROM SKIN AND CLOTHING WITH COLD WATER Background: Energy transfer is involved in almost all chemical and physical processes. One of the most common forms of energy transferred in a chemical or physical process is heat. Heat (g) is a form of energy (thermal energy) that can be transferred spontaneously from an object at a high temperature to an object at a lower temperature. When heat flows into a substance, the temperature of that substance will increase. The quantity of heat (g) required to cause a temperature change of any substance is related by the proportionality constant called the specific heat ($) of that substance. q = msAt The specific heat (s) is the amount of heat required to raise the temperature of one gram of the substance by 1°C. For example, the specific heat of water is 4.18 J/g· Thus, in order to raise the temperature of 1 g of water by 1°C, 4.18 joules of heat must be added to the water Although heat transfers cannot be measured directly, the changes in temperature that results from the heat transfer can. In this experiment, the amount of heat ( transferred from a warm sample of water to a cool sample of antifreeze will be calculated based on the change in temperature that occurs. From the amount of heat transferred, the specific heat of antifreeze will be calculated. 49

Explanation / Answer

1) MAss of water = Mass of water+ beaker -Mass of beaker

MAss of water = 155.27 - 63.62 = 91.65 grams

Change in temperature = Ti -Tf = 51 - 46 = 50C

2) The heat transferred by water = Q = Mass of water X specific heat of water X change in temperature

                                             = 91.65 X 4.18 X 5 = 1915.49 Joules

3) The heat transferred to antifreeze = Q = 1915.49 Joules

4) The change in temperature of antifreeze = Tf-Ti = 46 - 20.3 0C = 25.7 0C

The mass of antifreeze = 53.96 - 2.67 = 51.29 grams

5) Q = m X s X change in temperature

1915.49 Joules = 51.29 X specific heat of antifeeze X 25.7

Specific heat of antifreeze = 1.45 g / J 0C

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.