Question

When 1.00 kg lead (specific heat = 0.13 J g -1 C -) at 100.0 C is added to a quantity of water at 28.5C, the final temperature of the lead-water mixture is 35.2 C. What is the mass of water present?

Answer 1

3.21grams

Answer 2

sorry the answer is actually about 300grams

Answer 3

16.2g

Answer 4

How can we solve this question

Answer 5

[AI-generated answer — Explain Bot]

To find the mass of water present, we can use the concept of heat transfer. The heat gained by the water will be equal to the heat lost by the lead. The formula for heat transfer is:

Q = m * c * ΔT

Where:
Q is the heat transfer
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

First, let's find the heat lost by the lead:

Q_lead = m_lead * c_lead * ΔT_lead

Q_lead = (1.00 kg) * (0.13 J/g°C) * (100.0°C - 35.2°C)
Q_lead = 1.00 kg * 0.13 J/g°C * 64.8°C
Q_lead = 84.24 J

Now, let's find the heat gained by the water:

Q_water = m_water * c_water * ΔT_water

Since we are trying to find the mass of water, let's rearrange the formula:

m_water = Q_water / (c_water * ΔT_water)

To find Q_water, we can use the fact that Q_lead is equal to Q_water:

Q_water = Q_lead

Therefore:

m_water = Q_lead / (c_water * ΔT_water)

Plugging in the values:

m_water = 84.24 J / (4.18 J/g°C * (35.2°C - 28.5°C))
m_water = 84.24 J / (4.18 J/g°C * 6.7°C)
m_water ≈ 3.92 g

Therefore, the mass of water present is approximately 3.92 grams.