To treat a burn on your hand, you decide to place an ice cube on the burned skin. The mass of the ice cube is 16.1 g, and its initial temperature is -13.3 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.3 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.
***Question: I'd like to know if the equations for a similar question answered by Dr.bob222 is correct and did he use the correct equations.
Thanx.
I vaguely remember a similar question within the last 2-3 weeks but I remember I had a problem interpreting it because it wasn't clear, as this one is, about the water. I have a suggestion. Instead of trying to remember a problem of several weeks ago why don't we work this one? Do you have an answer. Post your work and we can go through it.
Please don't post questions piggy back on another. Go to the top of the page and post your own question. I am deleting this question and I will attempt to repost yours at the top of the page.
no
To determine if the equations used by Dr.bob222 are correct for the given scenario, we would need to know the specific equations used by him. However, I can walk you through the process of solving this problem using the appropriate equations.
To calculate the amount of heat absorbed by the ice cube and resulting water, we can use the equation:
Q = m * c * ΔT
Where:
Q is the heat absorbed (in Joules)
m is the mass of the ice cube (in grams)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (in °C)
First, we need to determine the change in temperature of the ice cube. The initial temperature of the ice cube is -13.3°C, and it melts and reaches the temperature of your skin, which is 31.3°C. Therefore, ΔT for the ice cube is:
ΔT = 31.3°C - (-13.3°C) = 44.6°C
Next, we calculate the heat absorbed by the ice cube using the equation:
Q_ice = m_ice * c_water * ΔT
Substituting the given values:
Q_ice = 16.1 g * 4.18 J/g°C * 44.6°C
Simplifying, we get:
Q_ice ≈ 3000 J
Next, we need to consider the water resulting from the melted ice. The heat absorbed by the resulting water will be equal to the heat released by the ice cube, as energy is conserved. Therefore, Q_water = -Q_ice (negative because heat is released).
So, Q_water = -3000 J
Therefore, the heat absorbed by the ice cube and resulting water is approximately 3000 J, assuming no heat is lost to the surroundings.