A 1.2 kg block of ice is initially at a temperature of -5°C.
(a) If 5.8 multiplied by 105 J of heat are added to the ice, what is the final temperature of the system?
°C
(b) Suppose the amount of heat added to the ice block is increased by a factor of 3.0. By what factor must the mass of the ice be increased if the system is to have the same final temperature?
multiplied by
You must do the (a) part in stages.
q1 = heat required to move the ice from a T of -5 to zero.
mass x specific heat ice x (Tfinal-Tinitial). You have mass, look up specific heat ice and delta T = 5.
q2 = heat to melt the ice.
q2 = mass x heat fusion.
You have mass. Look up heat of fusion.
Now q1 + q2 have been used. Subtract from the 5.8E5 J to determine the amount of heat remaining (if any remains--perhaps not all of the ice will melt) to raise the T of water from zero to unknown T.
q = massH2O x specific heat H2O x (Tfinal-Tinitial) and solve for Tfinal. Tinitial is zero. Post your work if you need further assistance.
Q1= 1.2 x 2.05 x 5 = 12.3
q2= 1.2 x 333 = 399.6
q1 + q2 = 411.9
580000 - 411.9 = 579588.1
what did i do wrong?
To solve this problem, we need to use the specific heat capacity of ice, which is 2.09 J/g°C, and the fact that the latent heat of fusion for ice is 334 J/g.
(a) To find the final temperature of the system, we can use the equation:
Q = mcΔT
Where:
Q = heat added to the ice (in Joules)
m = mass of the ice (in kg)
c = specific heat capacity of ice (in J/g°C)
ΔT = change in temperature (in °C)
To convert the mass of the ice from kg to g, we multiply by 1000:
m = 1.2 kg * 1000 g/kg = 1200 g
Plugging in the values:
5.8 * 10^5 J = 1200 g * 2.09 J/g°C * ΔT
ΔT = (5.8 * 10^5 J) / (1200 g * 2.09 J/g°C)
ΔT ≈ 224.64 °C
The final temperature of the system is -5°C + 224.64°C = 219.64 °C.
(b) If we increase the amount of heat added by a factor of 3, we need to find the new mass of the ice block to keep the final temperature the same.
Let's assume the new mass of the ice block is m'.
To keep the final temperature the same, we can set up the equation:
Q' = m'cΔT
Where:
Q' = new heat added to the ice (in Joules)
c = specific heat capacity of ice (in J/g°C)
ΔT = change in temperature (in °C)
Since the final temperature is the same, ΔT remains unchanged.
Therefore, the equation becomes:
Q' = m'cΔT = 3(5.8 * 10^5 J)
m'cΔT = 3(1200 g * 2.09 J/g°C * ΔT)
m' = (3(5.8 * 10^5 J)) / (1200 g * 2.09 J/g°C)
m' ≈ 7.83 kg
Therefore, the mass of the ice needs to be increased by a factor of approximately 7.83 / 1.2 = 6.53 to maintain the same final temperature.
To answer these questions, we need to use the concept of specific heat capacity and the formula for heat transfer.
(a) First, we need to calculate the heat required to raise the temperature of the ice block to its final temperature.
The formula for heat transfer is given by:
Q = m * c * ΔT
Where:
Q is the heat transfer (in Joules),
m is the mass of the object (in kilograms),
c is the specific heat capacity (in J/(kg·°C)),
ΔT is the change in temperature (in °C).
For ice, the specific heat capacity is approximately 2090 J/(kg·°C).
Given:
m = 1.2 kg (mass of the ice block)
c = 2090 J/(kg·°C) (specific heat capacity of ice)
ΔT = final temperature - initial temperature = final temperature - (-5°C) = final temperature + 5°C
We also know that Q = 5.8 * 10^5 J (given heat transfer).
Substituting the known values into the formula:
5.8 * 10^5 J = 1.2 kg * 2090 J/(kg·°C) * (final temperature + 5°C)
Now let's solve for the final temperature:
5.8 * 10^5 J = (1.2 kg * 2090 J/(kg·°C)) * (final temperature + 5°C)
Divide both sides by (1.2 kg * 2090 J/(kg·°C)):
(final temperature + 5°C) = (5.8 * 10^5 J) / (1.2 kg * 2090 J/(kg·°C))
(final temperature + 5°C) = 23.79°C
Subtract 5°C from both sides:
final temperature = 23.79°C - 5°C
(final temperature) = 18.79°C
So, the final temperature of the ice block is 18.79°C.
(b) Next, we need to determine the factor by which the mass of the ice block needs to be increased if the same final temperature is to be achieved with an increased heat transfer.
Let's call the increased heat transfer factor "F".
We know the mass of the ice block is initially 1.2 kg.
To achieve the same final temperature, the new mass of the ice block can be calculated using the formula for heat transfer:
Q = F * m_new * c * ΔT
where m_new is the new mass of the ice block.
Since the final temperature and specific heat capacity remain constant, the equation becomes:
F * m_new = Q / (c * ΔT)
We know that F = 3.0 (given increase in heat transfer), and we can substitute the other known values:
3.0 * m_new = (5.8 * 10^5 J) / (1.2 kg * 2090 J/(kg·°C) * 18.79°C)
Simplify the equation:
m_new = (5.8 * 10^5 J) / (1.2 kg * 2090 J/(kg·°C) * 18.79°C * 3.0)
m_new = 1.6066416 kg
Therefore, the mass of the ice block must be increased by a factor of 1.6066416 / 1.2 ≈ 1.339 for the system to have the same final temperature.
To summarize:
(a) The final temperature of the ice block is approximately 18.79°C.
(b) The mass of the ice block must be increased by a factor of approximately 1.339 for the system to have the same final temperature when the heat transfer is increased by a factor of 3.0.