A 20 gm piece of metal is heated to 150 c and then dropped into an aluminum calorimeter of mass 500gm containing 500gm of water initially at 25 c find the final equilibrium temperature of the system if the specific heat of the metal is 128.100j/kg-k while the specific heat of water is4200j/kg-k
Heat lost = Heat gained
"m_hs_h\u0394T_h = m_cs_c\u0394T_c \\\\\n\nm_h = 0.02 \\;kg \\\\\n\nm_c = 0.5 \\;kg \\\\\n\n0.02 \\times 128.1 (150 -T_f) = 0.5 \\times 4200 (T_f -25) \\\\\n\n384.3 -2.562T_f = 2100T_f -52500 \\\\\n\n52884.3 = 2102.562T_f \\\\\n\nT_f = \\frac{52884.3}{2102.562} = 25.15 \\;C"
Answer: 25.15 °C
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