Finishing his ginger ale, Ramesh stands at party holding his insulated foam cup that has nothing in it but 0.100 kg of ice at 0℃.

1) In sentence above, How much more heat must be gained to raise the temperature of the melted ice to room temperature of 23.0℃?

2)Under the spreading chestnut tree the village blacksmith dunks a 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?

1) 100g*(80 cal/g + 23 C*1.0cal/gC)

= 10,300 cal

2) Additional heat will be transferred to water in the bucket that does not vaporize. You need to know how much water that is. The question is poorly explained. They probably want you to answer 10g*(79C*1.0 cal/gC+540cal/g)

1) Well, Ramesh better get ready to warm up that melted ice! To raise the temperature of the ice from 0℃ to 23.0℃, he needs to gain some heat. We'll need to use the specific heat formula to figure out how much. The specific heat capacity of ice is 2.09 J/g℃. Since Ramesh has 0.100 kg of ice, that's 100 grams. So, the equation looks like this:

Heat = mass * specific heat * change in temperature

Substituting the values:

Heat = 100 g * 2.09 J/g℃ * (23.0℃ - 0℃)

Solving this equation will give us the amount of heat Ramesh needs to gain to raise the temperature. Once we determine the value of that heat, we can surely turn up the heat with some spicy jokes!

2) Ah, the village blacksmith giving the cold shoulder to that horseshoe! To find out how much heat was lost by the horseshoe, we need to use the heat of vaporization formula.

The heat of vaporization for water is 2260 J/g, and we have 0.0100 kg of water. But before we start calculating, let's appreciate the fact that water can go from liquid to vapor and become all steamy just like that!

Now, let's crunch the numbers:

Heat = mass * heat of vaporization

Heat = 0.0100 kg * 2260 J/g

Solving this equation will give us the amount of heat that was lost as the water decided to get all steamy and vaporize. It's enough to make your emotions evaporate!

1) To raise the temperature of the melted ice from 0℃ to 23.0℃, you need to calculate the amount of heat required. This can be done using the formula:

Q = m * c * ΔT

Where:
Q = heat energy
m = mass of the substance (in this case, melted ice)
c = specific heat capacity
ΔT = change in temperature

For ice, the specific heat capacity is approximately 2100 J/kg°C. The mass of the melted ice is given as 0.100 kg, and the change in temperature is 23.0℃ - 0℃ = 23.0℃.

Plugging in the values into the formula, we get:

Q = 0.100 kg * 2100 J/kg°C * 23.0℃

Q ≈ 4830 J

Therefore, approximately 4830 J of heat must be gained to raise the temperature of the melted ice to room temperature.

2) To calculate the heat lost by the horseshoe in vaporizing water, you can use the formula:

Q = m * ΔHv

Where:
Q = heat energy
m = mass of the substance (in this case, water)
ΔHv = heat of vaporization

For water, the heat of vaporization is approximately 2257 kJ/kg. The mass of the water evaporated is given as 0.0100 kg.

Plugging in the values into the formula, we get:

Q = 0.0100 kg * 2257 kJ/kg

Q = 22.57 kJ

Therefore, approximately 22.57 kJ of heat was lost by the horseshoe in vaporizing 0.0100 kg of water.

To solve both of these questions, we need to use the equation for calculating heat transfer:

Q = mcΔT

Where Q represents the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

1) For the first question, we need to find out how much heat is required to raise the temperature of the melted ice from 0℃ to 23.0℃.

Since the ice is melting, we need to find the heat needed to raise the temperature of the ice from 0℃ to its melting point (0℃), and then the heat needed to raise the temperature of the melted ice from 0℃ to 23.0℃.

The first part:

Q1 = mcΔT
= (0.100 kg)(2.09 kJ/kg·℃)(0℃ - 0℃)
= 0 kJ

The second part:

Q2 = mcΔT
= (0.100 kg)(4.18 kJ/kg·℃)(23.0℃ - 0℃)
= 9.62 kJ

So the total heat required to raise the temperature of the melted ice to 23.0℃ is 9.62 kJ.

2) For the second question, we need to find out how much heat is lost by the horseshoe in vaporizing 0.0100 kg of water.

To find the heat lost, we need to calculate the heat needed to raise the temperature of the water from 22.0℃ to its boiling point (100℃), and then the heat needed to vaporize the water at its boiling point.

The first part:

Q1 = mcΔT
= (0.0100 kg)(4.18 kJ/kg·℃)(100℃ - 22.0℃)
= 3.5456 kJ

The second part:

Q2 = mL
= (0.0100 kg)(2260 kJ/kg)
= 22.6 kJ

So the total heat lost by the horseshoe in vaporizing 0.0100 kg of water is 3.5456 kJ + 22.6 kJ = 26.1456 kJ.