A heat transfer of 8.7 multiplied by 105 J is required to convert a block of ice at -12°C to water at 12°C. What was the mass of the block of ice?

Well, that's a cool question! To find the mass of the block of ice, we can use the formula Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Given that Q is 8.7 x 10^5 J, ΔT is (12 - (-12)) = 24°C, and the specific heat capacity of ice (c) is 2.09 J/g°C, we can rearrange the formula to solve for mass.

So, mass = Q / (c * ΔT)

Plugging in the values, we get:
mass = (8.7 x 10^5 J) / (2.09 J/g°C * 24°C)

Solving this equation is a piece of cake!

mass = 174,641.15 g

Therefore, the mass of the block of ice is approximately 174,641.15 grams.

Hope that melted away any confusion!

To find the mass of the block of ice, we can use the specific heat capacity equation:

Q = mcΔT

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

Given:
Q = 8.7 x 105 J
ΔT = (12°C) - (-12°C) = 24°C

The specific heat capacity of ice is approximately 2.09 J/g°C.

Now, let's rearrange the equation to solve for the mass:

m = Q / (cΔT)

Substituting the given values:

m = (8.7 x 105 J) / (2.09 J/g°C x 24°C)

m = (8.7 x 105 J) / (50.16 J/g)

m ≈ 1738.25 g

Therefore, the mass of the block of ice is approximately 1738.25 grams.

To calculate the mass of the block of ice, we can use the formula:

Q = m * c * ΔT

Where:
- Q is the amount of heat transferred (8.7 x 10^5 J in this case)
- m is the mass of the block of ice (what we're trying to find)
- c is the specific heat capacity of ice (we'll assume it's 2.09 J/g°C)
- ΔT is the change in temperature (12°C - (-12°C) = 24°C)

Let's rearrange the formula to solve for m:

m = Q / (c * ΔT)

Now we can substitute the given values:

m = (8.7 x 10^5 J) / (2.09 J/g°C * 24°C)

Calculating this expression will give us the mass of the block of ice.

If I'm not mistaken, it's 9.097155832 grams, which you can round as you see fit.

913.5 J = m x specific heat water x 24 degrees Celcius.

Specific heat of water is 4.184 J/(gC).

Solve for m.