I have a few problems im having trouble with, any help..?

1.
If a peice of aluminum with mass 3.90g and a temperature of 99.3C is dropped into 10.0cm^3 of water at 22.6C, what will be the final temperature of the system?

2.
If a peice of cadmium with a mass 65.6g and a temp. of 100.0C is dropped into 25.0 cm^3 of water at 23.0C, what will be the final temp.

3.a piece of unknown metal with mass 23.8g is heated to 100.0C and dropped into 50.0cm^3 of water at 24.0.The final temp is 32.5C. What is the specific heat.

Help, these are soo confusing!!

how do i know the specific heat?

Number one is about 28.5 C

Sure, I can help you with these problems. The key concept you need to understand here is the principle of heat transfer, specifically the equation for calculating the heat gained or lost in a system. The equation 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 joules/gram°C)
- ΔT is the change in temperature (in °C)

Now, let's solve each problem step by step:

1. For the aluminum and water mixture:

To find the final temperature, we can assume no heat is lost to the surroundings (i.e., it's an isolated system).
Now, using the equation q = m * c * ΔT, we need to calculate the heat gained by aluminum and the heat lost by water, and set them equal to each other.

Let's calculate the heat gained by aluminum:
q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
q_aluminum = (3.90g) * (specific heat capacity of aluminum) * (final temperature - initial temperature)
Here, we need to look up the specific heat capacity of aluminum, which is 0.897 J/g°C.

Next, let's calculate the heat lost by water:
q_water = m_water * c_water * ΔT_water
q_water = (10.0cm^3) * (density of water) * (specific heat capacity of water) * (final temperature - initial temperature)
Here, we need to use the density of water, which is approximately 1g/cm^3, and the specific heat capacity of water, which is 4.18 J/g°C.

Since no heat is lost to the surroundings, q_aluminum = q_water. Therefore, we can set up the equation:
(3.90g) * (0.897 J/g°C) * (final temperature - 99.3°C) = (10.0g) * (1g/cm^3) * (4.18 J/g°C) * (final temperature - 22.6°C)

Solving this equation will give you the final temperature of the system.

2. For the cadmium and water mixture:

Follow the same steps as in problem 1, but this time use the properties specific to cadmium (density, specific heat capacity) and the given values to calculate the final temperature.

3. For the unknown metal and water mixture:

In this case, you need to calculate the specific heat capacity of the unknown metal.

Using the same equation q = m * c * ΔT, we can set up the equation:
(23.8g) * (specific heat capacity of unknown metal) * (final temperature - 100.0°C) = (50.0g) * (4.18 J/g°C) * (final temperature - 24.0°C)

Solving this equation will give you the specific heat capacity of the unknown metal.

Remember to look up the specific heat capacities and densities of the metals and water. These values may vary depending on the reference source.

I hope this explanation helps you solve these problems!

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

The above will solve any of these problems. Just plug in the numbers and solve for the one unknown.

I'm surprised they aren't given in the problem. At any rate, look up the specific heat of Al from problem 1 and Cd from problem 2. Problem 3 gives you temperatures and you are to calculate specific heat. The specific heat of water is 4.184 Joules/grams or 1.0 calories/gram.