Question

An electric heater raises the temperature of 120g of water in a thin light vessel through 10 K in 2 min., when placed in 70 g of water contained in a metal vessel of mass 0.55 kg the temperature rises through 9 K in the same time. Calculate from the above the heat supplied in 2 min by the heater b. the power of the heater a. C. the heat supplied to the 70 g of water d the heat supplied to the metal vessel the heat capacity of the vessel e. f. the specific heat capacity of its material (specific heat capacity of water = 4200 J.kg-¹k-¹)​

Answer 1

Notice in this question, there are two sets of containers with water.

the material of container one is thin so it was ignored.

The mass of water in container one is 120g=0.12kg(1000g=1kg)

TEMPERATURE rise=10k

time=2mins(2x60)=120s

For second container the material has a mass of 0.55kg.

and a mass of water 70g=70/1000=0.07kg.

Temperature rise= 9k

Note that since the heater is put in each of them for the same amount of time.

The heat supplied is equal for both.

see below.

Step-by-step explanation:

a. H=mc°

H=0.12 ×4200×10

H=5040J

b. P=H÷t (t= 2mins= 2×60s= 120secs)

P= 5040÷120

P=42 watts

c. H=mc°

H= 0.07×4200×9

H= 2646J

d.heat supplied by the heater is the same for both masses of water

heat supplied/heat lost= heat gained by 0.07kg of water + heat gained by metal vessel.

5040J= 2646J +H

H= 5040-2646=2394J

e. Heat capacity of the vessel cp= mc

H=cp°; cp=H÷° = 2394÷9= 266Jk^-1

f. mc= cp

c= cp÷m

c= 266÷0.55 = 483.6Jkg^-1k^-1.