A coffee-cup calorimeter contains 150.0 g of water at 25.3 degreesCelsius. A 124

Question

A coffee-cup calorimeter contains 150.0 g of water at 25.3 degreesCelsius. A 124 -g block of copper is heated at 100.4 degreesCelsius by putting it into a beaker of boiling water. The specificheat of Cu(s) is 0.385 J/g K. The Cu is added to the calorimeter,and after a time the contents of the cup reach a constanttemperature of of 30.1 degrees Celsius.

Question 1) Determine the heat in J, lost by copper(Cu) block.
Question 2) Determine the amount of heat gained by the water. Thespecific heat of the water is 4.18 J/g K.
Question 3) What would be the final temperature of the system ifall the heat of copper block were absorbed by the water in thecalorimeter.

Explanation / Answer

we know that heat lost by copper is equal to the heat gainedby water. so here, heat lost by water = mcT where m is mass            cis specific heat capacity            T is the temperature change we know that specific heat capacity of water is 4.18J/gK a)heat lost by copper = 124 *0.385 *(100.4 - 30.1)                              =3356.122 J b)so here heat gained by water = 150 *4.18 *(30.1 - 25.3) a)heat lost by copper = 124 *0.385 *(100.4 - 30.1)                              =3356.122 J                                             = 3009.6 J c)if all the heat of copper block were absorbed by the waterin the calorimeter. then in that case let the final temperature be t then heat gained by water = 150 *4.18 *(t - 25.3) heat lost by copper = 124 *0.385 *(100.4 - t) both are equal so, 124 *0.385 *(100.4 - 30.1) =150 *4.18 *(t - 25.3) solving this we get t = 30.65 degrees celcius (ans)

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