Esperimens k Caloeimetry Pre-Lab Questions Name: Section: L. Esplain the princip

Question

Esperimens k Caloeimetry Pre-Lab Questions Name: Section: L. Esplain the principle of calorimsctry that enabkes the mxcasanement of heat chunges 1. Explain the 2 objects wil 2 objedts will Reach the sameeu Terium Hep the in·tially cold object wil now be warmer and the wond an in latot have a verytughot av rylowtpecific leader, whe a ven heatenergy is absorbed at 750 "C? Show all work.a2g8 sas s. How much energy (in Joules) is needed to heat an iron nail wich a mass of 7.0 g fooes 25 " until it becomes red bot at 750 °Cz Show all wodk AT 750 -S725 72S' 4. Calculate the amount of energy (in Calories)'released daring the combastion of a peanut ehat heats up 100) g of water from 20 to 65 (Note: Food Calories penni!ckers where l Calorie = 1 kcal 1,00 cal) Show all work. q4.S ealerie 94.5 ealeri 4: 18828 18 8 18 416 73

Explanation / Answer

Q3)

Specific heat of Iron = 0.450 J/gC

Energy needed = mass * specific heat of iron * change in temperature

=> 7.0 g * 0.450 J/gC * (750-25)

=> 7.0 * 0.450 * 725

=> 2283.75 J

Q4)

Change in Temperature = 65-20 = 45

Specific heat of water = 4.179 J/gC

Amount of energy released = 100g * 4.179 J/gC * 45

=> 100 * 4.179 * 45

=> 18805.5J

1 Joule = 4.184 Calories

Hence ernergy in calories = 18805.5/4.184 = 4494.62 Cal

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