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Transcript
Heat Lost Heat Gained problems The heat lost by one substance in a system is gained by another in the system when there is a difference in temperature between the substances. There is also a heat transfer between the system and its surroundings if they are at different temperatures. Example: 100g of iron with a specific heat of 0.444J/g°C at 25°C is added to 50g of water at 80°C. Find the final temperature. Qlost=Qgained gCpΔT= gCpΔT (50g)(4.18J/g°C )(80°C-x)=(100g)( 0.444J/g°C)(x-25°C) For ΔT we have to do some substitution. We don’t know the final temperature and so we can’t directly calculate the change in temperature. Therefore we choose x to be the variable that represents the final temperature. We still need to plug in change in temperature to the equation. We want the change in temperature to be a positive value. We know that the heat will transfer from the higher temperature substance to the lower temperature substance. The higher temperature will lose heat and the lower temperature substance will gain heat. So 80 °C is a larger temperature than the final temp x. So a positive change in temperature can be made by setting up the change in temperature to be 80°C-x. The lower temperature substance (iron) will gain heat. Therefore the final temperature x is a larger value than 25°C. The change in temperature can be written as x-25°C for the iron. Simplify the multiplication, distribute, combine like terms and solve for x. (209)(80-x)=44.4(x-25) 16720-209x=44.4x-1110 16830=253.4x x=66.42°C Example 2: If 80g of gold at 50°C is added to 10g of water at 25°C, the final temperature rests at 30°C. Find the specific heat of gold in J/g°C. gCpΔT= gCpΔT (80g)(Cp)(20°C)=(10g)(4.18J/g°C)(5°C) Cp=0.13J/g°C