They mixed 0.2 kg of hot water with a temperature of 100 degrees C and cold water with a temperature of 25 degrees C. The temperature of the mixture was set at 40 degrees C. How much cold water was required for this? specific heat capacity of water is 4200j / kg
mg = 0.2 kg.
tg = 100 ° C.
tx = 25 ° C.
t = 40 ° C.
С = 4200 J / kg * ° C.
When mixing water of different temperatures, hot water will give the amount of heat to cold water: Qg = Qx.
The amount of heat Qg, which the hot water will give during cooling, is expressed by the formula: Qg = C * mg * (tg – t).
The amount of heat Qx that goes to heat cold water can be expressed by the formula Qx = C * mx * (t – tx).
C * mg * (tg – t) = C * mx * (t – tx).
mg * (tg – t) = mх * (t – tх).
The mass of cold water mx will be determined by the formula: mx = mg * (tg – t) / (t – tx).
mх = 0.2 kg * (100 ° C – 40 ° C) / (40 ° C – 25 ° C) = 0.8 kg.
Answer: when mixing cold water, it was mx = 0.8 kg.
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