Calculate the final temperature that results from mixing 245 grams of cobalt, specific heat 0.446 J/g0C, at 1420C with 106 grams of water at 24.80C.

bobpursley is incorrect

To calculate the final temperature when mixing two substances of different temperatures, we can use the equation:

(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0

Where:
m1 = mass of substance 1 (cobalt) in grams
m2 = mass of substance 2 (water) in grams
c1 = specific heat of substance 1 in J/g°C
c2 = specific heat of substance 2 in J/g°C
ΔT1 = change in temperature of substance 1
ΔT2 = change in temperature of substance 2
0 = final temperature (equilibrium temperature)

First, let's calculate the change in temperature for each substance:

ΔT1 = final temperature - initial temperature for cobalt
ΔT2 = final temperature - initial temperature for water

The initial temperature for cobalt is 1420°C, and the initial temperature for water is 24.8°C.

Now, we can plug in the values and solve for the final temperature:

(245g * 0.446 J/g°C * (final temperature - 1420°C)) + (106g * 4.186 J/g°C * (final temperature - 24.8°C)) = 0

(109.270J/°C * (final temperature - 1420°C)) + (443.116J/°C * (final temperature - 24.8°C)) = 0

Simplifying the equation:

109.270J/°C * final temperature - 15482.2340J = -443.116J/°C * final temperature + 10957.2208J

Combining like terms:

109.270J/°C * final temperature + 443.116J/°C * final temperature = 10957.2208J + 15482.2340J

552.386J/°C * final temperature = 26439.4548J

Finally, solve for the final temperature (°C):

final temperature = 26439.4548J / 552.386J/°C

final temperature ≈ 47.8°C

To calculate the final temperature that results from mixing the cobalt and water, we can use the principle of heat transfer. The heat gained by the water must be equal to the heat lost by the cobalt.

The formula to calculate the heat gained or lost is:
Q = m * c * ΔT

Where:
Q is the heat gained or lost (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)

For the cobalt:
Q1 = m1 * c1 * ΔT1

Where:
m1 = 245 grams (mass of cobalt)
c1 = 0.446 J/g°C (specific heat of cobalt)
ΔT1 = final temperature - initial temperature for cobalt = Tf - 1420°C

For the water:
Q2 = m2 * c2 * ΔT2

Where:
m2 = 106 grams (mass of water)
c2 = 4.184 J/g°C (specific heat of water)
ΔT2 = final temperature - initial temperature for water = Tf - 24.80°C

Since the heat gained by the water must be equal to the heat lost by cobalt, we can set up the equation:
Q1 = Q2

m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Now let's substitute the given values into the equation:

245 grams * 0.446 J/g°C * (Tf - 1420°C) = 106 grams * 4.184 J/g°C * (Tf - 24.80°C)

Now, let's simplify and solve for Tf:

109.07 (Tf - 1420) = 443.104 (Tf - 24.80)

109.07 Tf - 154867 = 443.104 Tf - 11009.93

333.034 Tf = 143857.07

Tf ≈ 431.88°C

Therefore, the final temperature after mixing is approximately 431.88°C.

heat needed to raise water to 100C:

.106*4.18*(100-24.8)=.331Kj
heat needed to change water to steam: 2264.76kJ/kg*.106= 240kj
Heat needed to raise steam from 100 to Tf.
.106*2kj/kg*(Tf-100)=???

Now, heat given off by cooling Cobalt:
.245*.446*(tf-1460)
sume of heats = 0
.245*.446*(tf-1460)-240-.106*2kj/kg*(Tf-100)-.331=0
solve for Tf check my work. Tf has to be above 100C, or the steam assumption was wrong.