Question

10 gm of steam at 100 degree celcius is mixed with 50 gm of ice at 0 degree celcius then final temperature is....???

Answer 1

The final temperature is 40°C.

Explanation:

10 gm steam to 10 gm water at 1000 °C

Heat released = 10 × 540 = 5400 cal

                        = 10 × 540 = 5400 cal

50 gm Ice to 50 gm water at 0°C

Heat absorbed = 50 × 80 = 4000 cal

• Let the final temperature be t°C.

So 10 gm water at 1000 C to 10 gm water at ​t°C

Heat released

= 10 × 1 × (100−t) = 1000^−10 t cal

50 gm water at 00 C to 50 gm water at ​t0 C

• Heat released

= 50 × 1 × (t−0) = 50t cal  = 50 × 1 × t-0 = 50 t cal

• Now

5400 + 1000 − 10t = 4000 + 50t

⇒ 60 x t = 2400

⇒t = 40°C

Thus the final temperature is 40°C.

Also learn more

If the pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C it's initial temperature must be??

cbse.brainly.in/strive-for-better/IN/1/

Answer 2

Answer:

16.66° c

Explanation:

The given datas are

Steam

mass (m₁) = 10 g

Temperature (T₁) = 100° C

ICE

mass (m₂) = 50 g

Temperature ( T₂) = 0° C

Let the final temperature be T₀

We know that , Q = m c dt              ( c = specific heat constant )

Heat lost by a body = heat gained by another body

m₁ c ( T₁ - T₀) = m₂ c ( T₀ - T₂)

10 * ( 100 - T₀) = 50 ( T₀ - 0⁰)

100⁰ - T₀ =  5T₀

T₀ = 16.66⁰ C

So the final temperature is 16.66° c