# HEY GUYS HELP ME???1 kg of ice at -10? is mixed with 4.4kg of water at 30? , the final temperature of mixture is...??

$We\;know,\\*Specific\;heat\;of\;water,s_{1}=4184\;\frac{J}{Kg\;K}\\* Specific\; heat\;of\;ice=2010\;\frac{J}{Kg\;K}\\* mass\;of\;water,m_{1} =4.4 \;Kg\\*mass \;of\;ice,m_{2}=1\;Kg\\* Temperature\;of\; ice,T_{2}=-10^{\circ} =263\;K\\*Temperature\;of\;water,T_{1}=30^{\circ}C=303\;K\\*latent\;heat\;of\;water,L= 2260\;\frac{KJ}{Kg}\\*Let\;the\;final\;temperature\;be,T\\* Here,\;Heat\;lost\;by\;4.4\;Kg\;water=Heat\;energy\;needed\;to\;change\;the\\*temperature\;of\;ice\;from\;-10^{\circ}C\;to\;0^{\circ}C+Latent\;heat\;needed\;to\;change\\*ice\;at\;0^{\circ}C\;into\;water\;at\;0^{\circ}C+heat\;absorbed\;by\;water\; (melted ice) \\*\Rightarrow m_{1}s_{1}(T_{1}-T)=m_{2}s_{2}(0-T_{2}) +m_{2}L+m_{2} s_{2}(T-T_{2})\\* \Rightarrow (4.4)(4184) (303-T)=(1)(2010)(0-(-263))+(1)(2260000)+(1)(2010)(T-263)\\* On\;solving\;above\;equation\;we\;get,\\* T=162.5\;K\\*So,\;final\;temperature\;is\;162.5\;K$

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