# Calculate the mass of water at 20 degrees C needed to lower the temperature of 750g of water at 75 degrees C to body temperature 37 degrees C?

neela | High School Teacher | (Level 3) Valedictorian

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We use the principle of heat loss  is equal to heat gain to solve the problem.

We are required to know the mass of water at  20 degrees  which reduces the temperature of 750g of water at 75 degree to 37 degree. Let the required mass of water be m gram.

So the heat gain by m gram of water when it cools 750g of water = mass *temperature increse = m(37-20) =17m calories.

Heat lost by the 750 gram of water  when it cools down from 75 deg C to 37 deg C = 750*(75-37) calories = 750*38 calories.

So, the heat gained by m grams of water = heat lost by 750gram of water. Or

17m cals heat gained by m gram of water = 750*38 cals of heat lost by 750 gram of water .  Solving for m, we get:

m = 750*38/17 gram. = 1676.47 gram of water is the mass of water at 20 degree C.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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Given that:

750 g (m1) of water at 75 degrees C (t1) is mixed with m2 g of water at 20 degrees C (t2). This result in the total mixture of water attaining a temperature of 37 degrees C (t).

We have to find out the value of m2.

The weight of total mixture = m1 + m2

The total heat required to heat a given mass of water to a given temperature is proportional to its mass multiplied by temperature.

Thus heat in a given mass of water =

H x Mass x Temperature.

Where H = specific heat of water.

Also total heat in mixture of the two initial quantities of water is equal to the sum of heat in initial quantities of water.

Thus:

H x (m1 + m2) x t = (H x m1 x t1) + (H x m2 x t2)

Dividing all terms of the equation by H we get:

(m1 + m2) x t = (m1 x t1) + (m2 x t2)

substituting values of m1, t1, t2, and t in the equation we get:

(750 + m2) x 37 = 750x75 + m2x20

2775 + 37m2 =  56250 + 20m2

37m2 - 20m2 = 56250 - 2775

17m2 = 53475

Therefore:

m2 = 53475/17 = 3145.5882 (approximately)