2022-04-29

An ice-cube tray of negligible mass contains 2.5 kg of water at 20°C. How much heat must be removed to cool the water to 0°C and freeze it?

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Skilled2022-05-14Added 403 answers

${m}_{water}=2.5kg$

$\u25b3T=0-20=-20\xb0C$

${c}_{water}=4190J/kg\xb0C$

${L}_{f.water}=3.34*{10}^{5}J/kg$

${Q}_{gain}+{Q}_{loss}=0$

${Q}_{gain}+{m}_{water}*{c}_{water}*\u25b3T-{m}_{water}*{L}_{f.water}=0$

${Q}_{gain}=-{m}_{water}*{c}_{water}*deltT+{m}_{water}*{L}_{f.water}=-2.5*4190*(-20)*2.5*3.34*{10}^{5}$

$=1.749325*{10}^{11}J$

$=165804018,4Btu$

22+64

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