2. One way to cool a cup of coffee would be to plunge an ice-cold piece of alumi

Question

2. One way to cool a cup of coffee would be to plunge an ice-cold piece of aluminum into it. Su 20.00 g piece of aluminum is stored in the refrigerator at 0.0°C and then dropped into a cup The coffee's temperature drops from 90.0°C to 75°C. How many kJ of thermal energy did t aluminum absorb? (Specific heat 0.902 /g.°C.) 1353 J or 1.353 Using information from question #2 above, calculate the mass (g) of coffee in the cup. (Specific heatoffe pecific heater 4.18 J/g. C.) 3. Between 0°C and 30°C, mercury has a specific heat of 0.138 J/g.°C. If 200 J of heat an 100 g of Hg at 20°C, what will the final temperature of the Hg be? (Hint: this is an e

Explanation / Answer

2. Heat absorbed by aluminium :

Mass of aluminium * Specific heat capacity of aluminium * (Final temperature - Initial Temperature)

= 20.0 g * 0.902 J / g 0C * ( 75 - 0 ) 0 C = 1353 Joules

3. Heat absorbed by aluminium = heat released by coffee.

Heat released by coffee = Mass of coffee * Specific heat capacity of aluminium * (Final temperature - Initial Temperature) = Mass of coffee * 4.18 J / g 0C * ( 75-90) 0C = - 62.7 * Mass of coffee J

( negative sign indicates that heat is relesed)

So, 62.7 * Mass of coffee = 1353

Mass of coffee = 1353/62.7 g = 21.6 g

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