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A 0.345 kg mass of tungsten at 130.0c is placed in a 0. 502 kg of water at 22.0C the the mixture reaches equilibrium at 28.6C calculate the specific heat of tungsten (specific heat of water = 4180 j/kg C).
Formula MaCa∆Tb = MbC∆Tb

Respuesta :

Answer: [tex]395.88J/kg^0C[/tex]

Explanation: Heat lost will be equal to the heat gained.

[tex]M_tC_t\Delta T_t=M_wC_w\Delta T_w[/tex]

where [tex]M_t[/tex]= mass of tungsten = 0.345 kg

[tex]C_t[/tex] = specific heat of tungsten

[tex]\Delta T_t[/tex] = Change in temperature of tungsten =[tex](130.0^0C-28.6^0C)=101.4^0C [/tex]

[tex]M_w[/tex]= mass of water = 0.502 kg

[tex]C_t[/tex] = specific heat of water = [tex]4180J/kg^0C[/tex]

[tex]\Delta T_w[/tex] = Change in temperature of water =[tex](28.6^0C-22.0^0C)=6.6^0C [/tex]

[tex]0.345\times C_t\times 101.4=0.502\times 4180\times 6.6[/tex]

[tex]C_t= 395.88J/kg^0C[/tex]




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