How much energy is required to convert 12.00 g of ice at -5.00 °C to liquid water at 2.50 °C?
Converting 12.0g12.0g12.0g of ice to 0°C0°C0°C
=12.0×10=1200calories=12.0\times 10=1200calories=12.0×10=1200calories
Converting to water at 2.5°C2.5°C2.5°C
=12.0g×2.5=30calories=12.0g\times 2.5=30 calories=12.0g×2.5=30calories
Latent heat of vaporization of water =537calories/g=537calories/g=537calories/g
=537×12=6444calories=537\times 12=6444 calories=537×12=6444calories
Total heat of fission=(1200+30+6444)=7674calories=(1200+30+6444)=7674 calories=(1200+30+6444)=7674calories
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