Question

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

Answer 1

Answer:
`395.88J/kg^0C`

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

`M_tC_t\Delta T_t=M_wC_w\Delta T_w`

where
`M_t`= mass of tungsten = 0.345 kg

`C_t` = specific heat of tungsten

`\Delta T_t` = Change in temperature of tungsten =
`(130.0^0C-28.6^0C)=101.4^0C`

`M_w`= mass of water = 0.502 kg

`C_t` = specific heat of water =
`4180J/kg^0C`

`\Delta T_w` = Change in temperature of water =
`(28.6^0C-22.0^0C)=6.6^0C`

`0.345* C_t* 101.4=0.502* 4180* 6.6`

`C_t= 395.88J/kg^0C`