45.0g of Silver (0.23 J/goc) was mixed in a beaker with 18.0g of water (4.18 J/g°C) with an initial temperature of 22.0°C. The final temperature was found to be 51.0°C. What was the change in temperature of the silver?

Answer: 210.8 °C

Explanation:

dа ta:

Silver:

m = 45.0g

C = 0.23 J / (g°C)

ΔT = ?

Water:

m = 18.0g

C = 4.18 J / (g°C=

Ti = 22.0°C

Equilibrium

Tf = 51.0°C

Solution:

1) At the end the system reaches the thermal equilibrium. That means that at the end all the system is at same temperature.

The problem states the the the final temperature is 51°C. That means that the sylver, the water and the beaker will be at 51°C

2) The heat released by the silver is the product of its mass, its heat capaciy and the change in temperature:

=> Heat released by silver = m * C * ΔT = 45.0g * 0.23 J / (g °C) * ΔT

3) The heat absorbed by the water is the product of its mass, its heat capacity and the change in temperature:

=> Heat absorbed by water = m * C * ΔT = 18.0g * 4.18 J / (g°C) * [51.0 - 22.0]°C

4) By the law of convervation of enery (first law of thermodynamics), assumin no losses of heat, the heat released by silver equals the heat absorbed by the water:

=> 45.0g * 0.23 J / (g °C) * ΔT = 18.0g * 4.18 J / (g°C) * [51.0 - 22.0]°C

From which you can solve for ΔT:

5) ΔT = 18.0g * 4.18J / (g°C) * (29°C) / (45.0g * 0.23J/g°C) = 210.8°

That is the answer, the change in temperature of silve was 210.8°C