A coffee-cup calorimeter contains 140.0 g of water at 25.3 degree C. A 124.0-g b

Question

A coffee-cup calorimeter contains 140.0 g of water at 25.3 degree C. A 124.0-g block of copper metal is heated to 100.4 degree C by pulling ii in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g middot K. The Cu is added to the calorimeter and after a time the contents of the cup reach a constant temperature of 30.1 degree C. Part B Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g middot K. Part C The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 calorimeter in J/K. Part D What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

Explanation / Answer

PART (D)

we know that, heat change,

q = m * s * (t2-t1)

If net heat change = 0

m * s * (t2-t1)water + m * s * (t2-t1)Cu = 0

124.0 * 4.184 * (t2 - 25.3) + 140.0 * 0.385 * (t2 - 100.4) = 0

518.816 t2 - 13126.045 + 53.9t2 - 5411.56 = 0

572.716 t2 = 18537.605

t2 = 32.37 0C

T = 273.15 + 32.37 K

T = 305.52 K

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