You need a water bath at 45 °C; if you start with 500 mlat 20 °C, how much 90 °C

Question

You need a water bath at 45 °C; if you start with 500 mlat 20 °C, how much 90 °C water must you add to achieve45 °C? (Assume no loss of heat to the surroundings,the density of water 1 g/ml,
Cpcontainer = 24 J/°C, Specific heat of water =4.2J/g °C). Remember that heat gained = - heat lost;do not forget the container. Will rate highand fast! You need a water bath at 45 °C; if you start with 500 mlat 20 °C, how much 90 °C water must you add to achieve45 °C? (Assume no loss of heat to the surroundings,the density of water 1 g/ml,
Cpcontainer = 24 J/°C, Specific heat of water =4.2J/g °C). Remember that heat gained = - heat lost;do not forget the container. Will rate highand fast!

Explanation / Answer

Remember that heat gained = - heat lost; do not forget thecontainer. now we get specific heat of water*temp change* water in grams + heat capacityof container*temp change = specific heat of water*temp change*water in grams(x) => 4.2*(45-20)*500 + 24*(45-20) =4.2*(90-45)*x    (where x is the grams of water at90 C) =>x = 280.95 g so as density of water = 1 g/l so volume of water needed to achieve 45 degrees celcius = 280.95ml

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