Respuesta :
Answer:
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
[tex]E = mc\delta T[/tex]
For water and for iron at equilibrium
[tex] m_{w}c_{w}\delta T=m_{i}c_{i}\delta T[/tex]
Put the value into the formula
[tex]22.54 \times0.449(T-21.34)=12.8\times4.184(98.14-T)[/tex]
[tex]10.12046(T-21.34)=53.5552(98.14-T)[/tex]
[tex]10.12046T-215.9706164=5255.907328-53.5552T[/tex]
[tex]10.12046T+53.5552T=5255.907328+215.9706164[/tex]
[tex]63.67566T=5471.8779444[/tex]
[tex]T=\dfrac{5471.8779444}{63.67566}[/tex]
[tex]T=85.93^{\circ}C[/tex]
Hence, The temperature will the metal and the water reach thermal equilibrium is 85.93°C
Answer: The final temperature of the system is 25.75°C
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.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of iron = 12.8 g
[tex]m_2[/tex] = mass of water = 22.54 g
[tex]T_{final}[/tex] = final temperature = ?°C
[tex]T_1[/tex] = initial temperature of iron = 98.14°C
[tex]T_2[/tex] = initial temperature of water = 21.34°C
[tex]c_1[/tex] = specific heat of iron = 0.449 J/g°C
[tex]c_2[/tex] = specific heat of water= 4.184 J/g°C
Putting values in equation 1, we get:
[tex]12.8\times 0.449\times (T_{final}-98.14)=-[22.54\times 4.184\times (T_{final}-21.34)][/tex]
[tex]T_{final}=25.75^oC[/tex]
Hence, the final temperature of the system is 25.75°C
