A 275-g sample of nickel at 100.0°C is placed in 100.0 g of water at 22.0°C. What is the final temperature of the water? Assume no heat transfer with the surroundings. The specific heat of nickel is 0.444 J/g·°C and the specific heat of water is 4.184 J/g·°C. Hint: Remember the final temp for both the system and surroundings will be the same, the definition of ΔT and the fact that ΔT of metal will not be same as water.

A. 39.6 °C

B. 40.8 °C

C. 79.2 °C

D. 82.4 °C

The final temperature of water (and nickel) is 39.6 °C

Explanation:

Step 1: Data given

Mass of the sample of nickel = 275.0 grams

Mass of water = 100.0 grams

Initial temperature of nickel = 100.0 °C

Initial temperature of water = 22.0°C

The specific heat of nickel is 0.444 J/g°C

The specific heat of water is 4.184 J/g°C.

Step 2: Calculate the final temperature

heat lost = heat gained

Qnickel = - Qwater

Q=m*c*ΔT

m (nickel) * c (nickel) * ΔT (nickel) = - m (water) * c (water) * ΔT (water)

⇒ the mass of nickel = 275.0 grams

⇒ c (nickel) = the specific heat of nickel = 0.444 J/g°C

⇒ ΔT (nickel) = The change of temperature = T2 - T1 = T2 - 100°C

⇒ the mass of water = 100.0 grams

⇒ c (water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT (water) = The change of temperature = T2 - 22.0°C

275.0g * 0.444 J/g°C * (T2-100°C) = - 100.0g * 4.184 J/g°C * (T2 - 22.0°C)

122.1 * (T2-100°C) = - 418.4 * (T2 - 22.0°C)

122.1T2 - 12210 = - 418.4T2 + 9204.8

122.1T2 + 418.4T2 = 9204.8 + 12210

540.5T2 = 21414.8

T2 = 39.6 °C

The final temperature of water (and nickel) is 39.6 °C