Solving Heat (calorimetry) transfer problems Finding the amount of heat to raise or lower the temp of a substance.

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Solving Heat (calorimetry) transfer problems Finding the amount of heat to raise or lower the temp of a substance

Heat Problems Q = mc(Tf-Ti) = Heat required to change the temperature of a material. c = specific heat and m = mass (kg) Q = mc(Tf-Ti) = Heat required to change the temperature of a material. c = specific heat and m = mass (kg) Q = heat, measured in joules Q = heat, measured in joules C(water) = 4186 J/kg k C(water) = 4186 J/kg k C(ice) = 2090 J/kg k C(ice) = 2090 J/kg k C (aluminum) = 900 J/kg k. C (aluminum) = 900 J/kg k.

Heat Problems When a material goes through a phase change (melting from solid to liquid, or boiling from liquid to gas) the amount of heat energy required is called the Latent heat. When a material goes through a phase change (melting from solid to liquid, or boiling from liquid to gas) the amount of heat energy required is called the Latent heat. L = latent heat L = latent heat L(melting water) = 334,000 J/kg L(melting water) = 334,000 J/kg L(boiling water) = 2,257,000 J/kg L(boiling water) = 2,257,000 J/kg Q = mL = heat required for a phase change Q = mL = heat required for a phase change

Heat problems Calculate the heat required to raise 2kg of water and aluminum from 25C to 60C Calculate the heat required to raise 2kg of water and aluminum from 25C to 60C Q = mc (Tf-Ti) Q = mc (Tf-Ti) Q(water) = (2kg) (4186 J/kg k) (60-25) = J Q(water) = (2kg) (4186 J/kg k) (60-25) = J Q(alum) = (2kg) (900 J/kg k) (60-25) = 63,000J Q(alum) = (2kg) (900 J/kg k) (60-25) = 63,000J

Heat Problems Calculate the heat required to bring 2 kg of ice at -20C to water at +5C Calculate the heat required to bring 2 kg of ice at -20C to water at +5C Q = Q(heat ice from -20C to 0C) + Q (melt the ice to water) + Q(heat water from 0 to 5C) Q = Q(heat ice from -20C to 0C) + Q (melt the ice to water) + Q(heat water from 0 to 5C) Q = mc(ice)(0- -20C) + mL(melting) + mc(water) (5 – 0C) Q = mc(ice)(0- -20C) + mL(melting) + mc(water) (5 – 0C) Q = 2kg(2090)(20C) + 2kg(334000J/kg k) + 2kg(4186)(5C) = J Q = 2kg(2090)(20C) + 2kg(334000J/kg k) + 2kg(4186)(5C) = J