A 155.0 g piece of copper at 179 oC is dropped into 250.0 g of water at 28.1 oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

The final temperature is 36.24 °C

Explanation:

Step 1: Data given

Mass of copper = 155.0 grams

Initial temperature of copper = 179.0 °C

Mass of water = 250.0 grams

The initial temperature of water = 28.1 °C

The specific heat of copper = 0.385 J/g°C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the final temperature

Qlost = -Qgained

Q =m*c*ΔT

Qcopper = -Qwater

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

⇒with mass of copper = 155.0 grams

⇒with c(copper) = the specific heat of copper = 0.385 J/g°C

⇒with ΔT = the change of temperature of copper = T2 - T1 = T2 - 179.0°C

⇒with mass water = 250.0 grams

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

⇒with ΔT = with ΔT = the change of temperature of copper = T2 - T1 = T2 - 28.1 °C

155.0 * 0.385 * (T2 - 179.0°C) = - 250 * 4.184 *  (T2 - 28.1 °C)

T2 = 36.24 °C

The final temperature is 36.24 °C

a is the answer to this question

explanation: