Question #23181

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Answer 1 · 2016-11-15T01:00:54

$"70.1 g"$ of water.

Ok, so I did a quick search and found the heat capacity for Cu.

The formula for thermal energy exchange is:

$Q_"released" + Q_"absorbed" = 0$

$(mcDeltaT)_ "released" + (mcDeltaT)_ "absorbed" = 0$

Where

$Q$ is the quantity of heat

$m$ is the mass in kg,

$c$ is the specific heat capacity

$DeltaT$ is the change in temperature, $(T_2 - T_1)$ .

Released is the Copper and the absorbed is the water.

$(mcDeltaT)_ "Cu" + (mcDeltaT)_ "water" = 0$

$0.02(380)(22-100) + m(4180)(22-20) = 0$

$-592.8 + 8360m = 0$

$8360m = 592.8 implies m = 0.071$

Since

$"0.071 kg " = " 70.1 g"$

you have $"70.1 g"$ of water.

Hope this helps :)