Jetc34
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Two samples of water are mixed together.
The first sample has a mass of 0.50 kg and is at 0°C. The second sample has a mass of 1.5 kg and is at 100°C. What is the equilibrium temperature of the water, assume this is a closed system? (Cwater = 4.18 kJ/kg*°C.)
A.75°C
B.67°C
C.33°C
D.50°C

Respuesta :

jcherry99

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x)            Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x)                          Remove the brackets.

0.5x = 150 - 1.5x                                  Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x             Combine like terms  

2x = 150                                               Divide by 2

x = 75

Answer

A

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