Question

cold water= 22.7805g
hot water= 63.4954g
$\Delta T$ hot= -20.78 Celcius
$\Delta T$ cold= 4.71 Celcius

Determine the energy released or absorbed by the hot and cold water.

Answer 1

cold water = 22.7805g

hot water = 63.4954g

$\Delta T$ hot = -20.78 Celcius

$\Delta T$ cold = 4.71 Celcius

Qhot = m Cp $\Delta T$ hot

           = 63.4954 x 4.18 x -20.78

           = - 5515 J

here energy released by hot water = - 5515 J

Q cold = m Cp $\Delta T$ cold

             = 22.7805 x 4.18 x 4.71

             = 448.5 J

here energy absorbed by cold water = 448.5 J