A 110. G of sample of copper (copper = 0.385 J/g·°C) is heated to 82.4°C and then placed in an insulated container of water (water = 4.18 J/g·°C) at 22.3°C. The final temperature of the water and copper is 24.9°C. What is the volume of water? (water = 1 g/mL)

Volume of water is 224mL

Explanation:

The heat that sample of copper reducing its temperature is the same heat that is absorbed from the water using the equation:

C(Cu)*m*ΔT = -C(water)*m*ΔT

Where C is specific heat of Copper and water, m is the mass and ΔT is change in temperature.

We need to solve the mass of water and as density of water is 1g/mL the mass in grams is equal to volume in mL

The copper change from 82.4°C to 24.9°C = 57.5°C

Mass of copper = 110g

Specific heat of copper and water is 0.385J/g°C and 4.18J/g°C, respectively.

Mass of water is our incognite and change in temperature of water is 22.3°C - 24.9°C = -2.6°C

Replacing:

C(Cu)*m*ΔT = -C(water)*m*ΔT

0.385J/g°C*110g*57.5°C = -4.18J/g°C*m*-2.6°C

224g = Mass of water

That means the volume of water is 224mL

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.