Question

A 12.8-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.

Answer 1

Answer: The final temperature of the system is 25.75°C

Step-by-step explanation:

When iron is dipped in water, the amount of heat released by iron will be equal to the amount of heat absorbed by water.

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of iron = 12.8 g

`m_2` = mass of water = 22.54 g

`T_(final)` = final temperature = ?°C

`T_1` = initial temperature of iron = 98.14°C

`T_2` = initial temperature of water = 21.34°C

`c_1` = specific heat of iron = 0.449 J/g°C

`c_2` = specific heat of water= 4.184 J/g°C

Putting values in equation 1, we get:

`12.8* 0.449* (T_(final)-98.14)=-[22.54* 4.184* (T_(final)-21.34)]`

`T_(final)=25.75^oC`

Hence, the final temperature of the system is 25.75°C

Answer 2

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

Step-by-step 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

`E = mc\delta T`

For water and for iron at equilibrium

`m_(w)c_(w)\delta T=m_(i)c_(i)\delta T`

Put the value into the formula

`22.54 *0.449(T-21.34)=12.8*4.184(98.14-T)`

`10.12046(T-21.34)=53.5552(98.14-T)`

`10.12046T-215.9706164=5255.907328-53.5552T`

`10.12046T+53.5552T=5255.907328+215.9706164`

`63.67566T=5471.8779444`

`T=(5471.8779444)/(63.67566)`

`T=85.93^(\circ)C`

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