Hot Steel; Cold Ice: A 36.0 g block of ice sits at −50.0 °C under 1.00 atm constant pressure in a perfectly insulated container. A 225 g mass of stainless steel at 125 °C is dropped onto the ice, and the system is closed. When the system reaches thermal equilibrium, ice remains. What is the value of q_steel? Enter your answer in Celsius using no decimal places
Note that:
delta H fus (H2o) = 333 j/g
delta H vap (H2o) = 2260 J/g
Cp (steel) = 0.500 J g^-1°C^-1
S(H2o(s)) = 2.11 J g^-1 °C^-1
S (H2o(f)) =4.18J g^-1 °C^-1
S (H2o(g)) = 2.00 J g^-1 °C^-1
Heat is in Joules, not degrees Celsius.
If some ice is still there then it is at 0 deg C
Anyway the steel is cooled from 125 C to 0 C
heat out of steel = 225 g (125 - 0) deg C * 0.500 Joules / g deg C
so would it be 14062?
That is what I get.
Remember it is heat out so you should call it negative.
To find the value of q_steel, we can use the equation:
q_steel = m_steel * Cp_steel * ΔT_steel
where:
m_steel = mass of stainless steel (225 g)
Cp_steel = specific heat capacity of stainless steel (0.500 J g^-1 °C^-1)
ΔT_steel = change in temperature of stainless steel
First, let's calculate the change in temperature of the stainless steel. Since the initial temperature is 125 °C and the final temperature will be the same as the temperature of the ice (-50.0 °C), we have:
ΔT_steel = (-50.0 °C) - (125 °C)
ΔT_steel = -175 °C
Note: In this calculation, we subtract the final temperature from the initial temperature since the temperature of the ice is lower than that of the stainless steel.
Now, we can substitute the values we know into the equation:
q_steel = (225 g) * (0.500 J g^-1 °C^-1) * (-175 °C)
Calculating this, we get the value of q_steel as:
q_steel = -19,375 J
Therefore, the value of q_steel is -19,375 J.
Note: The negative sign indicates that heat is being lost by the stainless steel during the process.