<span>6.50x10^3 calories.
Now we have 4 pieces of data and want a single result. The data is:
Mass: 100.0 g
Starting temperature: 25.0°C
Ending temperature: 31.5°C
Specific heat: 1.00 cal/(g*°C)
And we want a result with the unit "cal". Now you need to figure out what set of math operations will give you the desired result. Turns out this is quite simple. First, you need to remember that you can only add or subtract things that have the same units. You may multiply or divide data items with different units and the units can combine or cancel each other. So let's solve this:
Let's start with specific heat with the unit "cal/(g*°C)". The cal is what we want, but we'ld like to get rid of the "/(g*°C)" part. So let's multiply by the mass:
1.00 cal/(g*°C) * 100.0 g = 100.0 cal/°C
We now have a simpler unit of "cal/°C", so we're getting closer. Just need to cancel out the "/°C" part, which we can do with a multiplication. But we have 2 pieces of data using "°C". We can't multiply both of them, that would give us "cal*°C" which we don't want. But we need to use both pieces. And since we're interested in the temperature change, let's subtract them. So
31.5°C - 25.0°C = 6.5°C
So we have a 6.5°C change in temperature. Now let's multiply:
6.5°C * 100.0 cal/°C = 6500.0 cal
Since we only have 3 significant digits in our least precise piece of data, we need to round the result to 3 significant figures. 6500 only has 2 significant digits, and 6500. has 4. But we can use scientific notation to express the result as 6.50x10^3 which has the desired 3 digits of significance. So the result is 6.50x10^3 calories.
Just remember to pay attention to the units in the data you have. They will pretty much tell you exactly what to add, subtract, multiply, or divide.</span>
The answer is carbon, silicone, germanium
Answer : The final temperature is, 
Explanation :
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of water at 
= 150 g
= mass of water at 
= 100 g
= final temperature = ?
= temperature of lead = 
= temperature of water = 
= same (for water)
Now put all the given values in equation (1), we get
![150\times (T_{final}-363)=-[100\times (T_{final}-303)]](https://tex.z-dn.net/?f=150%5Ctimes%20%28T_%7Bfinal%7D-363%29%3D-%5B100%5Ctimes%20%28T_%7Bfinal%7D-303%29%5D)
Therefore, the final temperature is,
