A 440.0 g piece of copper is dropped into 400.0 mL of water at 20.00 °C. The final temperature of the water was measured at 35.80 °C. Calculate the initial temperature of the metal. (SCu = 0.3850 J/gC)

207.50°C

Explanation:

Mass of metal (M1) = 440g

Volume of water = 400mL

Initial temperature of water (T2) = 20°C

Final temperature of the mixture (T3) = 35.80°C

Specific heat capacity of the metal (C1) = 0.3850J/g°C

Specific heat capacity of water = 4.184J/g°C

Density of water = 1g/mL

Temperature of metal (T1) = ?

Heat lost by the metal = Heat gained by the water

Q = MC∇T

Q = heat energy

M = Mass

C = specific heat capacity

∇T = change in temperature

M1C1(T1 - T3) = M2C2(T3 - T2)

M2 = ?

Density = mass / volume

Mass = density * volume

Mass = 1g/mL * 400mL

Mass = 400g

M1C1(T1 - T3) = M2C2(T3 - T2)

400 * 0.3850 (T1 - 35.80) = 400 * 4.184 * (35.80 - 20)

154(T1 - 35.80) = 1673.6 * (15.8)

154T1 - 5513.2 = 26442.88

154T1 = 26442.88 + 5513.2

154T1 = 31956.08

T1 = 31956.08 / 154

T1 = 207.50

The initial temperature of the metal was 207.50°C

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

the answer is na (sodium)