Thermodynamics question?

Question

Your electricity has gone out and the water in the water heater is at 12 oC. If you run 95 Liters of water in the tub, how many Liters of water at 95 oC will you have to add to the tub to raise the temperature to 38 oC? Specific Weight of water is 9.81 KN/M3

Answer 1

I don't know where we can use your given specific weight of water of 9.81 kn/m^3. But for this problem we will use the volumetric specific heat of water which is K = 4.124 N/cm^3/degree Kelvin;

Now the Heat absorbed by a material is given by the expression;

Q = VK dT

Where:

Q = heat energy in joules

V = volume in liters (1000 cm^3)

K = specific heat in joules/cm^3/ degree Kelvin

dT =  change in temperature in degrees Kelvin

The heat absorbed by the 95 liters of water from 12 degrees C to 38 degrees centigrade is;

(1). Q = 95 x 1000 x K x (38 - 12) = 95 x 1000 x26K joules

Q is the same amount of heat released by the volume of water added which has a temperature change from 95 degrees C to 38 degrees C. Thus;

(2). Q = V x 1000 x K x (95 -38) = V x 1000 x 57K joules

Hence;

95 x 1000 x 26 K = V x 1000 x 57K

95 x 26 = 57 V

V = 95 x26/57 = 43.333 liters