Heat gained by the copper:
"q_1=m_cC_c\u0394T\n\n = (120) (0.40 ) (40 \u2212 0)\n\n= 1920\\ J" Heat gained by the water:
"q_2=m_wC_w\u0394T\n\n = (70) (4.2 ) (40 \u2212 0)\n\n= 11760\\ J" Heat gained by the ice:
"q_3 = m_iL + m_iC_w\u0394T\n\n\\\\ = (10) (320 ) + (10 ) (4.2 ) (40-0 )\n\n = 4880\\ J" Heat lost by the steam:
"q _4= mL_s + mC_w\u0394T\n\n \\\\= m (2200) + m (4.2 ) (100 \u2212 40)\n\n = 2452 m"
"1920 + 11760 + 4880 = 2452 m"
"m = 7.6\\ g"
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