1) Peter is heating water on the stove to boil eggs for a picnic. How much heat is required to raise the temperature of his 10.0-kg vat of water from 20.0℃ to 100.0℃?

Formula>> Q=mTcp

2)Under the spreading chestnut tree of the village blacksmith dunks red-hot horseshoe into a large bucket of 22.0℃ water. How much heat was lost by the horseshoe in vaporizing 0.0100 kg of water?
Formula>> 2.26x10^6J/kg

3)By January, the 3.0 kg of water in the birdbath in Robyn's backyard has frozen to a temperature of -7.0℃. As the season changes, how much heat must be added to a water to make it a comfortable 25℃ for the birds?
formula>> Q=mTcp

Please help me!! :)
Thank you soooo much!! :D

dwadaw

1.) is 3350 J

1) To find the amount of heat required to raise the temperature of the water, we can use the formula: Q = m * Cp * ΔT.

Given:
mass of water (m) = 10.0 kg
specific heat of water (Cp) = 4186 J/kg℃
temperature change (ΔT) = (100.0℃ - 20.0℃) = 80.0℃

Using the formula, we can calculate the heat required:
Q = 10.0 kg * 4186 J/kg℃ * 80.0℃

Q = 3,348,800 J

Therefore, the heat required to raise the temperature of the water is 3,348,800 Joules.

2) To find the amount of heat lost by the horseshoe in vaporizing water, we can use the formula: Q = m * Lv.

Given:
mass of water vaporized (m) = 0.0100 kg
latent heat of vaporization (Lv) = 2.26x10^6 J/kg

Using the formula, we can calculate the heat lost:
Q = 0.0100 kg * 2.26x10^6 J/kg

Q = 22,600 J

Therefore, the heat lost by the horseshoe in vaporizing water is 22,600 Joules.

3) To find the amount of heat required to raise the temperature of water from -7.0℃ to 25℃, we can use the formula: Q = m * Cp * ΔT.

Given:
mass of water (m) = 3.0 kg
specific heat of water (Cp) = 4186 J/kg℃
temperature change (ΔT) = (25℃ - (-7.0℃)) = 32℃

Using the formula, we can calculate the heat required:
Q = 3.0 kg * 4186 J/kg℃ * 32℃

Q = 401,472 J

Therefore, the heat required to raise the temperature of the water is 401,472 Joules.

You're welcome! I'm here to help. Let's go through each question and calculate the required values step by step.

1) To find the amount of heat required to raise the temperature of Peter's 10.0-kg vat of water from 20.0℃ to 100.0℃, we'll use the formula Q = m * Cp * ΔT.

Here's how to calculate it:
- Mass (m) of water = 10.0 kg
- Specific heat capacity (Cp) of water = 4.18 J/g℃ (or 4.18 J/gK)
- ΔT (change in temperature) = final temperature - initial temperature = 100.0℃ - 20.0℃ = 80.0℃

Now, let's substitute the values into the formula:
Q = 10.0 kg * 4.18 J/g℃ * 80.0℃

Convert mass to grams:
Q = 10,000 g * 4.18 J/g℃ * 80.0℃

Multiply all the values together and calculate:
Q = 3,344,000 J

Therefore, the amount of heat required is 3,344,000 Joules.

2) To determine the amount of heat lost by the horseshoe during the vaporization of 0.0100 kg of water, we'll use the latent heat of vaporization formula:

Q = m * Lv

Here's how to calculate it:
- Mass (m) of water vaporized = 0.0100 kg
- Latent heat of vaporization (Lv) of water = 2.26 x 10^6 J/kg

Now, let's substitute the values into the formula:
Q = 0.0100 kg * 2.26 x 10^6 J/kg

Calculate:
Q = 22,600 J

Therefore, the amount of heat lost is 22,600 Joules.

3) To find out how much heat must be added to the water to raise its temperature from -7.0℃ to 25.0℃, we'll use the same formula as in question 1:

Q = m * Cp * ΔT

Here's how to calculate it:
- Mass (m) of water = 3.0 kg
- Specific heat capacity (Cp) of water = 4.18 J/g℃ (or 4.18 J/gK)
- ΔT (change in temperature) = final temperature - initial temperature = 25.0℃ - (-7.0℃) = 32.0℃

Now, let's substitute the values into the formula:
Q = 3.0 kg * 4.18 J/g℃ * 32.0℃

Convert mass to grams:
Q = 3,000 g * 4.18 J/g℃ * 32.0℃

Multiply all the values together and calculate:
Q = 398,400 J

Therefore, the amount of heat required is 398,400 Joules.

I hope this helps! Let me know if you have any more questions.