Answer :
Answer:
30.63 °C will be the final temperature of the water.
Explanation:
Heat lost by iron will be equal to heat gained by the water
[tex]-Q_1=Q_2[/tex]
Mass of iron = [tex]m_1=20 g [/tex]
Specific heat capacity of iron = [tex]c_1=0.444 J/g^oC [/tex]
Initial temperature of the iron = [tex]T_1=120^oC[/tex]
Final temperature = [tex]T_2[/tex]=T
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Volume of water = 300 ml
Density of water = 1 g/mL
Mass of water= [tex]m_2=300 mL\times 1 g/mL = 300 g[/tex]
Specific heat capacity of water= [tex]c_2=4.184 J/g^oC [/tex]
Initial temperature of the water = [tex]T_3=30^oC[/tex]
Final temperature of water = [tex]T_2[/tex]=T
[tex]Q_2=m_2c_2\times (T-T_3)[/tex]
[tex]-Q_1=Q_2[/tex]
[tex]-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)[/tex]
On substituting all values:
we get, T = 30.63°C
30.63 °C will be the final temperature of the water.