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?

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)

Answer:

3145.5882 g of water