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