1100g of ice at -50C is mixed with 100g of steam at 120C. Find out how much ice melts? The implied final temperature is 0C.
Cice=0.5 Csteam=0.46 Hv=539 Hf =80
here is my work
1100(0.5)(0+50)+80m = 100(0.46)(120-0)+540(100)
i am not getting the right answer
First look at the Right side of your equation. The steam cools from 120 to 100, not to 0. Then that 100 g of condensed steam cools to 0 C with a specific heat of 1.00. The 540*100 term is OK.
On the Left side, the entire 1100 g of ice heats up from -50 to 0C with a specific heat of 0.50, and then a mass M melts with a latent heat of fusion of 80. The left side is OK. Solve for M.
All this assumes that you are left with icewater. If m turns out negative, or greater than 1100g, all of the ice may have melted, or remained below 0 C. In that case, you will have to rewrite the equation for that possibility.
44554
To find out how much ice melts, we need to calculate the heat transferred from the steam to the ice until they reach a common final temperature of 0°C.
Let's break down the calculation step by step:
1. Calculate the heat gained by the ice to reach 0°C:
heat gained by ice = mass of ice (m) * specific heat capacity of ice (Cice) * change in temperature
= 1100g * 0.5 * (0°C - (-50°C))
= 1100g * 0.5 * 50°C
= 27,500 calories (cal)
2. Calculate the heat lost by the steam to reach 0°C:
heat lost by steam = mass of steam (m) * specific heat capacity of steam (Csteam) * change in temperature
= 100g * 0.46 * (120°C - 0°C)
= 100g * 0.46 * 120°C
= 5,520 calories (cal)
3. Calculate the heat released by the steam during condensation:
heat released by steam = mass of steam (m) * latent heat of vaporization (Hv)
= 100g * 539 cal/g
= 53,900 calories (cal)
4. Calculate the heat absorbed by the melted ice during the phase change:
heat absorbed by melted ice = mass of melted ice (m) * latent heat of fusion (Hf)
= m * 80 cal/g
Since the final temperature is 0°C, the heat gained by the ice equals the sum of the heat lost by the steam and the heat released during condensation.
Now, let's form the equation:
27,500 + m * 80 = 5,520 + 53,900
Simplifying the equation:
27,500 + 80m = 59,420
Finally, solve for 'm' – the mass of melted ice:
80m = 59,420 - 27,500
m = (59,420 - 27,500) / 80
m ≈ 344 grams
Therefore, approximately 344 grams of ice melts when 1100 grams of ice at -50°C is mixed with 100 grams of steam at 120°C, and the final temperature is 0°C.