Question

800 calories of heat is required to raise the temperature of 0.080 18
liquid from 10 °C to 100 °C. Find its specific capacity in cal kg .
1) 111.1
2) 2.222
3) 11.11
4) 2222​

Answer 1

Answer:

2 ) 2.222

Step-by-step explanation:

option 2

i hope it helps u

Answer 2

Specific heat capacity of the liquid is 1) 111.1 cal

Mass of the liquid = M = 0.080 Kg (Given)

Heat supplied = Q = 800 calories. (Given)

Initial temperature = 10°C (Given)

Final Temperature = 100°C.(Given)

Therefore,

Temperature difference = 100° - 10° = 90°

Now,

Q = mcT, where c is the specific heat

c = Q / mΔT  

c = 800/ 0.080 × 90

c = 800 / 7.2 cal

c =  111.1

Thus, the specific heat capacity is 111.1 kg