You have 23.0g of water initially at -46°C. How much energy in joules is required to

heat the ice to 0°C? c ice = 2.1 J/g°C. I thought joules was calculated by grams*change in temperature*specific heat. The answer I get from 23*46*2.1 is 2221.8, but web assign says this is wrong, please help.

I calculated the same number as you; I suspect the problem is that you're reorting too many significnt figures. If you posted the problem EXACTLY as stated, then you're allowed only two places by the 2.1 and I would round the answer to 2.2E3 J. If that 2.1 is really 2.10 I would report the answer as 2.22E3

• A 50.0 g sample of ice at 0°C is melted. How much heat was absorbed by the ice?

To calculate the energy required to heat the ice to 0°C, you need to consider the specific heat capacity of ice and the change in temperature.

The equation you mentioned, Joules = grams * change in temperature * specific heat, is correct. However, there seems to be a minor mistake in the given values.

The specific heat capacity of ice (c_ice) is indeed 2.1 J/g°C, so you have that part correct.

You correctly have 23.0g of ice (grams) and the change in temperature is (0°C - (-46°C)) = 46°C.

So, the correct calculation would be:
Joules = grams * change in temperature * specific heat
Joules = 23.0g * 46°C * 2.1 J/g°C
Joules = 2232.6 J (rounded to one decimal place)

Hence, the correct answer is 2232.6 Joules, which may be slightly different from the value you obtained due to rounding differences.

To calculate the energy (in joules) required to heat the ice to 0°C, you are using the correct formula: energy = mass * change in temperature * specific heat.

However, there are a few factors that need to be considered in this problem:

1. The first factor is the phase change of water from solid (ice) to liquid (water) at 0°C. When water undergoes a phase change, it requires an additional amount of energy called the heat of fusion. In this case, you need to heat the ice from -46°C to 0°C, but you also need to account for the energy required to melt the ice at 0°C.

The heat of fusion for water is 334 J/g. This means that for every gram of ice, you need to provide an additional 334 joules of energy to change it from a solid to a liquid at its melting point.

So, the total energy required to heat the ice from -46°C to 0°C and melt it into liquid water at 0°C can be calculated as follows:

Energy = (mass of ice * change in temperature * specific heat of ice) + (mass of ice * heat of fusion)

2. Another important factor is that water at -46°C is already in the solid state. Hence, before using the formula mentioned above, you need to calculate the energy required to heat the ice from -46°C to its melting point (0°C).

So, the steps to solve this problem are as follows:

Step 1: Calculate the energy required to heat the ice from -46°C to 0°C:
Energy1 = (mass of ice * change in temperature * specific heat of ice)

Step 2: Calculate the energy required to melt the ice at 0°C:
Energy2 = (mass of ice * heat of fusion)

Step 3: Add both energies together to get the total energy required:
Total Energy = Energy1 + Energy2

Now, let's proceed with the calculations:

Step 1:
mass of ice = 23.0g
change in temperature = 0°C - (-46°C) = 46°C
specific heat of ice = 2.1 J/g°C

Energy1 = (23.0g * 46°C * 2.1 J/g°C)

Step 2:
mass of ice = 23.0g
heat of fusion = 334 J/g

Energy2 = (23.0g * 334 J/g)

Step 3:
Total Energy = Energy1 + Energy2

Calculate the values and sum them up to find the total energy required to heat the ice to 0°C.