Answer:49.3
Explanation:
Given
mass of water [tex]m_w=4 kg[/tex]
mass of tank [tex]m_T=13 kg[/tex]
initial of temperature of water =[tex]50^{\circ}C[/tex]
initial of temperature of tank=[tex]27^{\circ}C[/tex]
Specific heat of water =4.184kJ/kg k
Specific heat of copper=0.386 KJ/kg k
From first law of thermodynamics we have
[tex]Q=\Delta U+W[/tex]
Given tank is insulated thus Q=0
[tex]-W=\Delta U[/tex]
work will be negative as it is being done on system
[tex]-\left ( -100\right )=\Delta U_{water}+\Delta U_{tank}[/tex]
[tex]-\left ( -100\right )=m_w\times c_{water}\left ( T-50\right )+m_Tc_{tank}\left ( T-27\right )[/tex]
[tex]100=4\times 4.184\left ( T-50\right )+13\times 0.386\left ( T-27\right )[/tex]
[tex]T=49.3^{\circ} C[/tex]
