12-6. Neglect heat gained from or lost to the in this 40.0 g of H,O(s) at 0°Care added to a quantity of H2O() that has a tempera. surroundings ture of 60.0°C. The final temperature reached is 10.0°C. To what quantity of H,O() was the ice added?

Question

12-6. Neglect heat gained from or lost to the in this 40.0 g of H,O(s) at 0°Care added to a quantity of H2O() that has a tempera. surroundings ture of 60.0°C. The final temperature reached is 10.0°C. To what quantity of H,O() was the ice added?

I got an answer of 72 or 80 g. Which is the right answer with steps?

Answer 1

You need to proof your question and correct as needed.

Answer 2

well Dr.Bob,

I got m(1)(60-10)=40(80)+80(1)(100

What you think?

I keep messing up.

Answer 3

Neglect heat gained from or lost to the in this 40.0 g of H2O(s) at 0.)°C added to a quantity of H2O() that has a tempera. surroundings temp of 60.0°C. The final temperature reached is 10.0°C. To what quantity of H2O() was the ice added?

Answer 4

To find the correct answer, we need to use the principle of conservation of energy, which states that the heat lost by one substance is equal to the heat gained by another substance in a closed system.

Let's break down the problem and solve it step by step:

1. Determine the initial heat of the water at 60.0°C:
The specific heat capacity of water is approximately 4.18 J/g°C. Since we have no information about the actual mass of water initially present, let's assume it is "m" grams.
The initial heat can be calculated using the formula: Q = m * c * ΔT, where ΔT is the change in temperature.
Q1 = mcΔT1 = m * 4.18 J/g°C * (10.0°C - 60.0°C)
Q1 = m * 4.18 J/g°C * (-50.0°C)

2. Determine the heat gained by the ice:
When ice melts, it absorbs heat without a change in temperature. The heat of fusion for water is approximately 334 J/g.
The heat gained by the ice can be calculated using the formula: Q2 = m * H_fusion
Q2 = 40.0 g * 334 J/g

3. Calculate the final heat of the water at 10.0°C:
The final heat can be calculated using the formula: Q3 = mcΔT3, where ΔT is the change in temperature.
Q3 = m * 4.18 J/g°C * (10.0°C - 60.0°C)
Q3 = m * 4.18 J/g°C * (-50.0°C)

According to the principle of conservation of energy:
Q1 + Q2 = Q3

Now, let's solve for m (the mass of water initially present):
Q1 + Q2 = Q3
m * 4.18 J/g°C * (-50.0°C) + 40.0 g * 334 J/g = m * 4.18 J/g°C * (-50.0°C)

Simplifying the equation:
-209m + 13360 = -209m

After canceling out the -209m terms, we are left with:
13360 = 0

Since the equation doesn't yield a meaningful solution, it seems there might be an error in the information provided or the calculations. Double-check the given values and calculations to obtain the correct answer.