A student determined the calorimeter constant of the calorimeter. The student added 50.00 mL of cold water to 50.00 mK of heated , distilled water in a styrofoam cup. The initial temp of the cold water was 21,00C and the hot water, 29.15C. The maximum temp pf the mixture was found to be 24.81C. Assume the density of water is 1.00g mL-1 and the specific heats is 4.184 J g-1 K-1

1) Determine the temp change for the hot water and cold water.

[Mass cold water x specific herat x (Tfinal-Tinitial)] + [mass hot water x specific heat x (Tfinal-Tinitial)]

Calculate Tfinal.
Then subtract Tfinal-24.81 to obtain delta T.

To determine the temperature change for the hot and cold water, you need to calculate the difference between the initial and final temperatures.

Given:
Initial temperature of the cold water (T1) = 21.00°C
Initial temperature of the hot water (T2) = 29.15°C
Maximum temperature of the mixture (Tf) = 24.81°C

For the cold water:
Temperature change (ΔT1) = Tf - T1

For the hot water:
Temperature change (ΔT2) = Tf - T2

Let's calculate the temperature changes for both:

ΔT1 = 24.81°C - 21.00°C
ΔT1 = 3.81°C

ΔT2 = 24.81°C - 29.15°C
ΔT2 = -4.34°C

Note that the temperature change for the hot water is negative because it lost heat to the cold water during the mixing process.

To determine the temperature change for the hot water and cold water, you need to calculate the difference between the initial and final temperatures for each.

For the hot water:

Temperature change = final temperature - initial temperature
Temperature change = 24.81°C - 29.15°C

For the cold water:

Temperature change = final temperature - initial temperature
Temperature change = 24.81°C - 21.00°C

Now let's calculate the temperature changes:

For the hot water:
Temperature change = 24.81°C - 29.15°C
Temperature change = -4.34°C

For the cold water:
Temperature change = 24.81°C - 21.00°C
Temperature change = 3.81°C

So, the temperature change for the hot water is -4.34°C, and the temperature change for the cold water is 3.81°C.