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

A.75°C

Let the change in temp of cold water be  x degrees,

while that of  hot water be 100 - x degrees.

Heat exchange =  mcΔt

Ice

Δt = x

m = 0.50 kg

c = 4.18 kJ/kg*°C

Hot water

Δt = 100 - x

m = 1.5 kg

c = 4.18

But;

Heat lost = heat gained

0.50 * c * x = 1.5 * c * (100 - x)          

0.50 *x = 1.5*(100 - x)                          

0.5x = 150 - 1.5x                                  

0.5x + 1.5x = 150 - 1.5x + 1.5x            

2x = 150                                              

x = 75° C

Hence; the equilbrium temperature will be 75° C

answer: option c.

explanation: