Question

1) Let's consider an ice cube has the volume of 20 mL.
The density of ice is 0.9 g/mL. What is the mass of the ice cube?
My answer: 0.9 * 20= 18gr

We would now like to calculate the total energy that is required to heat up this piece of ice from -196˚C to body temperature, 37˚C. We know that for this to happen, the ice will transform to water at 0˚C. Thus we need to do this in 3 steps:

1: Raise the temperature of the ice from -196˚C to 0˚C.

2: Melt the ice, converting it from a solid at 0˚C to a liquid at 0˚C.

3: Raise the temperature of the water from 0˚C to 37˚C

1)Use the specific heat equation, 𝑄=𝑚𝑐𝑝Δ𝑇 , to determine how much energy it takes to heat the ice from -196˚C to 0˚C. Assume that the specific heat of ice is 2.11 J/g ⋅ K
My answer:18*2.11*(-196-0) = 7444 j/g

2)Determine how much energy it takes to melt the ice, using the latent heat of fusion (L = 334 J/g) in the equation 𝑄=𝑚𝐿
My answer:18*334=6012 j/g

3)Use the specific heat equation to determine how much energy it takes to heat the water from 0˚C to 37˚C?
My answer:18*4.186*37=2787 j/g

4)Now we can add these three numbers to calculate the total energy needed to bring the ice cube from -196˚C to body temperature. What is this total energy in Joules?
My answer: 7444+6012+2787= 16243 j/g

Answer 1

typo
18*2.11*(-196-0) = 7444 j/g
I think you mean
18*2.11*(0 - (-196)) = 7444 j/g
final temp - original temp

18*4.186*37=2787 j/g { I get 2788}
so I get 16244

{0.8 is 1, not 0 :}

Answer 2

Thank you Mr Damon ;)