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Physics 20 Online
OpenStudy (anonymous):

A 100-g ice cube at 0°C is placed in 650 g of water at 25°C. What is the final temperature of the mixture?

OpenStudy (anonymous):

you need to check Q's first. Energy you need to melt 100g of ice \[Q_1=mL=100g*80cal/gr=8000cal\] energy released when 650 gr of water goes from 25 to zero: \[Q_2=mc\Delta t=650*1*(25-0)=16.250cal\] \[Q_2>Q_1\] so this means 650 gr ow water will be able to melt thta ice and they will meet somewhere between 0 and 25C. lets say that temperature is x, then \[m_1L+m_1c(x-0)=m_2c(25-x)\] you need to solve this equation for x: 8000+100x=16250-650x 750x=8250 x=11C

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