Answered

The specific heat of iron is 0.11 cal /g·°C. What will be the final temperature if 275 calories are added to a 75.0 piece of iron initially at 25°C?

Answer :

Answer:

[tex]T_2=58.33\ ^0C[/tex]

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  

[tex]\Delta H[/tex]  is the enthalpy change

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]  is the temperature change

Thus, given that:-

Mass  = 75.0 g

Specific heat = 0.11 cal/g°C

[tex]\Delta T=T_2-T_1=T_2-25.0\ ^0C[/tex]

Heat added = 275 calories

So,  

[tex]275=75.0\times 0.11\times (T_2-25.0)=31.88713\ J[/tex]

Solving, we get that:-

[tex]T_2=58.33\ ^0C[/tex]

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