joules needed to melt 30.0 g of ice at 0 ∘C and to warm the liquid to 60.0 ∘C

so i did Q=mFf which i set up like
30.g x 334 j/c = 10020 J

then i'm sorta confused what i'm doing next. explain in detail please?

To find the total energy (in joules) required to melt 30.0 g of ice at 0°C and then warm the resulting liquid to 60.0°C, you need to consider two separate processes: the energy required for melting the ice and the energy required for raising the temperature of the resulting liquid.

1. Energy required to melt the ice:
The formula you used, Q = m × ΔHf, is correct. Q represents the heat energy (in joules), m is the mass of the substance (in grams), and ΔHf is the heat of fusion (or enthalpy of fusion) of the substance (in J/g).

For ice, the heat of fusion is approximately 334 J/g. So, plugging in the values:
Q = (30.0 g) × (334 J/g) = 10,020 J
This means it takes 10,020 joules of energy to melt the ice.

2. Energy required to warm the liquid:
To find the energy required to warm the liquid from 0°C to 60°C, you need to use the formula Q = m × c × ΔT. Here, c represents the specific heat capacity of the substance (in J/g°C), and ΔT represents the change in temperature (in °C).

The specific heat capacity of water is approximately 4.18 J/g°C. Plugging in the values:
Q = (30.0 g) × (4.18 J/g°C) × (60°C - 0°C) = 7,524 J
This means it takes 7,524 joules of energy to warm the liquid from 0°C to 60°C.

Now, to find the total energy required, you add the two values together:
Total energy = energy to melt ice + energy to warm liquid
Total energy = 10,020 J + 7,524 J = 17,544 J

Therefore, the total energy required to melt 30.0 g of ice at 0°C and warm the resulting liquid to 60.0°C is 17,544 joules.

To calculate the total energy required to melt the ice and warm the resulting liquid, you need to consider two separate steps: (1) melting the ice and (2) heating the liquid water.

Step 1: Melting the ice
To calculate the energy (Q) required to melt the ice, you correctly used the equation Q = m × ΔHf, where m is the mass of the substance and ΔHf is the heat of fusion, which is 334 J/g for ice. Plugging in the given values:
Q = 30.0 g × 334 J/g = 10,020 J

So, you correctly found that it takes 10,020 Joules to melt the ice.

Step 2: Heating the liquid water
To calculate the energy required to warm the liquid water from 0°C to 60°C, you need to consider the specific heat capacity (c) of water. The specific heat capacity represents how much energy is needed to raise the temperature of a substance by 1 degree Celsius per gram.

The specific heat capacity of water is approximately 4.18 J/g°C. Using the equation Q = m × c × ΔT, where ΔT is the change in temperature, we can calculate the energy required to warm the liquid water.

ΔT = 60°C - 0°C = 60°C

Plugging in the values for specific heat capacity (c = 4.18 J/g°C), mass (m = 30.0 g), and ΔT (60°C):
Q = 30.0 g × 4.18 J/g°C × 60°C = 7,524 J

So, it takes 7,524 Joules to heat the water from 0°C to 60°C.

To find the total energy required, you simply add the energy from each step:

Total Energy = Energy to melt ice + Energy to warm water
Total Energy = 10,020 J + 7,524 J = 17,544 J

Therefore, it takes a total of 17,544 Joules to melt 30.0 grams of ice at 0°C and warm the resulting liquid water to 60°C.

That melts the ice at zero to liquid water at zero C.

Now to heat it from zero C to 60 C is q = mc*delta T.
q = 30 x 4.184 x (Tfinal-Tinitial)
Then add the two for the total.