n ice bag containing 204 g of ice at 0 ∘C was used to treat sore muscles. When the bag was removed, the ice had melted and the liquid water had a temperature of 33.5 ∘
As importantly, what is your question?
To find the amount of heat absorbed by the ice, you can use the equation:
Q = m * c * ΔT
Where:
Q is the heat absorbed or released (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g⋅°C)
ΔT is the change in temperature (in °C)
In this case, we need to find the amount of heat absorbed by the ice, so the equation becomes:
Q = 204 g * c * (33.5 °C - 0 °C)
Now, we need to find the specific heat capacity of ice. The specific heat capacity of a substance is the amount of heat required to raise the temperature of a given mass of the substance by 1 degree Celsius.
The specific heat capacity of ice is 2.09 J/g⋅°C.
Substituting this value into the equation, we can calculate the heat absorbed by the ice:
Q = 204 g * 2.09 J/g⋅°C * (33.5 °C - 0 °C)
Calculating this expression will give us the amount of heat absorbed by the ice in Joules.