In this experiment a hot metal block at 100oC is added to cold water in a calorimeter. The heat lost by the metal block equals the heat gained by the water and the two end up at the same temperature.

In one experiment, the mass of the metal block is 89.5 grams, the volume of water used is 75.6 mL, the temperature of the cold water extrapolated to the time of mixing is 18.2oC, and the temperature of the metal block and water extrapolated back to the time of mixing is 28.3oC.

The specific heat of water is 4.184 J K-1g-1

How much heat is gained by the cold water?

Heat gained by cold water = ____ J

What is the specific heat of the metal?

Specific heat of the metal= ___J K-1g-1

According to the Law of Dulong and Petit, what is the molar mass of the metal?

Molar mass = ____ g mol-1

mass metal x specific heat metal x (Tfinal-Tinitial) + mass water x specific heat water x (Tfinal-Tinitial) = 0

You can pick out the pieces you want from above to answer your questions.

For Dulong and Petit.
heat capacity = about 24.9 J/mol*K. (about 6.3 calories/g*C

n one experiment, the mass of the metal block is 93.3 grams, the volume of water used is 96.3 mL, the temperature of the cold water extrapolated to the time of mixing is 20.4oC, and the temperature of the metal block and water extrapolated back to the time of mixing is 28.7oC.

The specific heat of water is 4.184 J K-1g-1

What is the molarity of ZnCl2 that forms when 25.0g of zinc completely reacts with CuCl2 according to the following reaction? Assume a final volume of 285mL .

Zn(s)+CuCl2(aq)→ZnCl2(aq)+Cu(s)

To find the heat gained by the cold water in this experiment, we can use the equation:

Q = m * c * ΔT

where
Q is the heat gained or lost,
m is the mass of the substance,
c is the specific heat of the substance, and
ΔT is the change in temperature.

For the cold water, we have:
- Mass (m) = 75.6 mL (volume) * 1 g/mL (density) = 75.6 g
- Specific heat (c) = 4.184 J K^(-1) g^(-1)
- Change in temperature (ΔT) = Temperature at end - Temperature at start = 28.3°C - 18.2°C = 10.1°C

Substituting these values into the formula, we can calculate the heat gained by the cold water:

Q = 75.6 g * 4.184 J K^(-1) g^(-1) * 10.1°C
Q = 3185.6784 J

Therefore, the heat gained by the cold water is approximately 3185.6784 J.

To find the specific heat of the metal, we can rearrange the formula:

Q = m * c * ΔT

to

c = Q / (m * ΔT)

Using the values from the experiment:
- Mass (m) of the metal block = 89.5 g
- Change in temperature (ΔT) = Temperature at end - Temperature at start = 28.3°C - 18.2°C = 10.1°C
- Heat gained by cold water (Q) = 3185.6784 J

Substituting these values into the formula, we can calculate the specific heat of the metal:

c = 3185.6784 J / (89.5 g * 10.1°C)
c = 3.5564 J K^(-1) g^(-1)

Therefore, the specific heat of the metal is approximately 3.5564 J K^(-1) g^(-1).

To find the molar mass of the metal using the Law of Dulong and Petit, we can use the formula:

Molar mass = (Specific heat * Atomic mass of the metal) / 3R

Where R is the gas constant, which is approximately 8.314 J K^(-1) mol^(-1) in SI units.

Rearranging the formula, we get:

Atomic mass of the metal = (Molar mass * 3R) / Specific heat

Substituting the values:
- Molar mass = unknown
- Specific heat = 3.5564 J K^(-1) g^(-1)
- R = 8.314 J K^(-1) mol^(-1)

We can calculate the atomic mass of the metal:
Atomic mass of the metal = (unknown * 3 * 8.314 J K^(-1) mol^(-1)) / 3.5564 J K^(-1) g^(-1)

The unit J cancels out, and we can simplify the equation to:

Atomic mass of the metal = 8.314 J K^(-1) mol^(-1) * unknown / 3.5564 J K^(-1) g^(-1)

Atomic mass of the metal = 23.46212217 * unknown

To solve for the unknown (molar mass of the metal), we need additional information or values.