Question Imagine that you have 25 kg of 25 degrees Celsius water. You
want to boil water on the stove and pour it into the rest of the water to raise
the temperature. Answer This is a pretty straightforward calorimetry problem. The first equation to be useful here calculates the amount of heat that you will need to raise the water up to temperature: q=ms∆t Plugging and chugging with your values for your starting conditions you get: q = (25000g)(4.184 J/gºC)(37ºC-25ºC) q = 1255200 J = 1255.2 KJ That gives you the amount of heat energy that you will need to raise your water to the desired temperature. Now, this value can be substituted in the above equation, along with the specific heat (constant) and a new delta T (equal to the difference in temperature between boiling water and the desired temperature of 37ºC) and solve it for the mass of water necessary to loose that much heat. Note that the heat energy is negative because it will be the heat that is given up by the water instead of absorbed like above. q=ms∆t m = -1255200 J / [(4.184 J/gºC)(37ºC-100ºC)] m = -1255200 J / (-263.592 J/g) m = 4762 g = 4.762 Kg So, we calculated the amount of heat that it would take to raise the water from room temp up to 37ºC and using that as a starting point we calculated the amount of boiling water that would be necessary to give up the energy necessary to do the desired task while not lowering itself below 37ºC. Final answer has been rounded to match significant figures in specific heat of water, assuming 0's trailing decimal place in two values given in problem.
Addendum - Question The heat lost by the boiling water is equal to the heat gained by the room-temperature water. How much heat was transferred in this process? Answer What you're asking for is just the value of 'q' in the question above in both instances when I used it. In the first equation that I solved, I was solving for how much energy needed to be gained by the water to raise it to the desired temperature (1255.2 KJ). In the second time the equation was used, I used this same value (1255.2 KJ) as the amount of heat energy that had to be lost by the boiling water to raise the 25 kg of water to temperature. Thus the answer you are looking for is 1255.2 KJ or 1255200 J.
Reference Chang, Raymond. Chemistry. 6th ed. USA, McGraw-Hill. 1998 |
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