A12.8-gram sample of iron is heated to 98.14°c in a boiling water bath. the hot metal is then immersed in 22.54 grams of water that is initially at 21.34°c. at what temperature will the metal and the water reach thermal equilibrium? the specific heat of iron is 0.449 j/g·°c. the specific heat of water is 4.184 j/g·°c.

The temperature will the metal and the water reach thermal equilibrium is 85.93°C

Explanation:

Given that,

Mass of iron = 12.8 gm

Temperature = 98.14°C

Mass of water = 22.54 gm

Temperature = 21.34°C

Specific heat of iron= 0.449 J/g°C

Specific heat of water= 4.184 J/g°C

Let final equilibrium temperature is T

We need to calculate the final temperature

Using formula of energy

For water and for iron at equilibrium

Put the value into the formula

Hence,  The temperature will the metal and the water reach thermal equilibrium is 85.93°C

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