Respuesta :
Answer:
30.0 grams is the mass of the steel bar.
Explanation:
Heat gained by steel bar will be equal to heat lost by the water
[tex]Q_1=-Q_2[/tex]
Mass of steel= [tex]m_1[/tex]
Specific heat capacity of steel = [tex]c_1=0.452 J/g^oC[/tex]
Initial temperature of the steel = [tex]T_1=2.00^oC[/tex]
Final temperature of the steel = [tex]T_2=T=21.40^oC[/tex]
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Mass of water= [tex]m_2= 105 g[/tex]
Specific heat capacity of water= [tex]c_2=4.18 J/g^oC [/tex]
Initial temperature of the water = [tex]T_3=22.00^oC[/tex]
Final temperature of water = [tex]T_2=T=21.40^oC[/tex]
[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:
[tex](m_1\times 0.452 J/g^oC\times (21.40^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.40^o-22.00^o))[/tex]
we get, [tex]m_1 = 30.0314 g\approx 30.0 g[/tex]
30.0 grams is the mass of the steel bar.
