When ice melts, it absorbs 0.33 per gram.How much ice is required to cool a 12.0 drink from 72 to 37, if the heat capacity of the drink is 4.18 ? (Assume that the heat transfer is 100 efficient.)
You didn't furnish any units on anything.
your dumb
He's right you didn't mention if the temperature from 72 to 37 is in Celsius or Fahrenheit?
How much ice is required to cool a 15.0 oz drink from 71 F to 34F, if the heat capacity of the drink is 4.18 J/g*C? (Assume that the heat transfer is 100 % efficient.)
To solve this problem, we need to calculate the amount of heat transferred from the drink to the ice.
First, let's calculate the amount of heat transferred when the drink cools from 72 to 37 degrees Celsius. We can use the equation:
Q = m * C * ΔT
Where:
Q is the amount of heat transferred
m is the mass of the drink
C is the heat capacity of the drink
ΔT is the change in temperature
The heat capacity of the drink is given as 4.18 J/g°C, the initial temperature is 72°C, and the final temperature is 37°C. Let's substitute these values into the equation:
Q = m * 4.18 * (37 - 72)
Next, since the heat transferred is equal to the heat absorbed by the ice, we can use the specific heat capacity of ice to find the mass of ice required. The specific heat capacity of ice is given as 0.33 J/g°C.
Let's rearrange the equation to solve for the mass of ice:
m_ice = Q / (0.33 * ΔT)
Substituting in the value of Q from above, and ΔT = 72 - 37 = 35°C, we can calculate the mass of ice required:
m_ice = (m * 4.18 * (37 - 72)) / (0.33 * 35)
Now we can substitute the values of m_ice, C, and ΔT into the equation to find the mass of ice required.
m_ice = (12.0 * 4.18 * (37 - 72)) / (0.33 * 35)
Simplifying this equation gives us the mass of ice required to cool the drink from 72 to 37 degrees Celsius.