Question

27. 3.5 kg of water (c=4189 J/(kgxK)) is heated from T1= 14 degrees to T2= 21.5 degrees.

1. Input an expression for the heat transferred to the water, Q.
2. Calculate the value of heat transferred to the water Q in joules, using the expression from part (a).
3. To the heated water 0.5 kg of water at T3= 20 degrees is added. What is the final temperature of the water, in kelvin

Answer 1

a ).

Q = mS

Q = Amount of heat supplied

m = mass of substance

S = Specific heat of substance

= rise in temperature = change in temperature of substance

Speicific heat of water ,S = 4189

Mass of water, m = 3.5 kg

Amount of heat supplied = 3.5 * 4189 * J

b ) .

= 21.5 - 14 = 7.5 K

Using formula for the heat transfer

Q = mS

Q = 3.5 * 4189 * 7.5

Q = 109961.25 J = 110 KJ

c ) .

Since 3.5kg of water is at higher temperature and the 0.5 kg of water is at lower temperature therefore heat will be transfer from 3.5 kg water to 0.5 kg water until they reach in equilibrium .

Let T be the equilibrium temperature .

Heat lost by the 3.5 kg water = heat gained by 0.5 kg water

3.5 * 4189 *(21.5 - T) = 0.5 * 4189 *(T -20)

7 * (21.5 - T) = T - 20

150.5 - 7T = T - 20

8T = 170.5

T = 21.3125 degree

T = 21.3125 + 273.16

T = 294.4725 K