(a) How much heat in kcal must be provided for a 0.853 kg block of lead to raise the
temperature of 0.015 kg block of ice from -5ºC to 0ºC? (b) How much heat in kcal must be
provided for a 0.853 kg block of lead to melt a 0.015 kg block of ice from at 0ºC? (c) How
much heat in kcal must be provided for a 0.853 kg block of lead to raise the temperature of
0.015 kg of liquid water from 0ºC to 5ºC? (d) What is the total amount of heat in kcal needed
for the whole process? (e) To what temperature must a 0.853 kg block of lead be heated in
order to complete this task? (specific heat for lead = 0.0305 kg/kcal*Cº)(Assume that there is
no heat lost to the surroundings.)
(a) You need the specific heat of ice, C(ice). Look it up.
Then use Q1 = M*C(ice)*deltaT
deltaT = 5C in this case
M = 0.015 kg
(b) You need the heat of fusion of ice, H. It is 80 kcal/kg
Then use
Q2 = M*H
(c) Use the specific heat of water,
C(liq)= 1.00 kcal/kg*C
Then use Q3 = M*Cliq*deltaT
deltaT = 5C in this case
(d) Add up Qtotal = Q1 + Q2 + Q3
(e) M(lead)*C(lead)*(T - 5) = Qtotal
They tell you the values of M(lead) and C(lead)
Solve for T
Thank you for your help!
You're welcome!
To solve these questions, we need to use the specific heat formula. The formula for heat (Q) is given by:
Q = m * c * ΔT
Where:
Q = heat (in kcal)
m = mass (in kg)
c = specific heat (in kcal/kg*Cº)
ΔT = change in temperature (in ºC)
Let's solve each part of the question step by step:
(a) How much heat in kcal must be provided for a 0.853 kg block of lead to raise the temperature of a 0.015 kg block of ice from -5ºC to 0ºC?
To solve this, we need to consider two separate processes:
1. Heating the ice from -5ºC to 0ºC (solid to solid phase change)
2. Heating the ice from 0ºC to the melting point (solid to liquid phase change)
1. Heating the ice from -5ºC to 0ºC:
Q1 = m1 * cice * ΔT1
Where:
m1 = mass of ice (in kg)
cice = specific heat of ice (in kcal/kg*Cº)
ΔT1 = change in temperature (in ºC)
Given:
m1 = 0.015 kg
cice = 2.09 kcal/kg*Cº
ΔT1 = 0ºC - (-5ºC) = 5ºC
Plugging in the values:
Q1 = 0.015 kg * 2.09 kcal/kg*Cº * 5ºC = 0.15675 kcal
2. Heating the ice from 0ºC to the melting point:
Q2 = m2 * cice * ΔT2
Where:
m2 = mass of ice (in kg)
ΔT2 = change in temperature (in ºC), which is the melting point of ice (0ºC)
Given:
m2 = 0.015 kg
ΔT2 = 0ºC
Plugging in the values:
Q2 = 0.015 kg * 2.09 kcal/kg*Cº * 0ºC = 0 kcal
Now, we can calculate the total heat provided for the whole process (a):
Q_total = Q1 + Q2 = 0.15675 kcal + 0 kcal = 0.15675 kcal
So, the total heat required to raise the temperature of the 0.015 kg block of ice from -5ºC to 0ºC is 0.15675 kcal.
(b) How much heat in kcal must be provided for a 0.853 kg block of lead to melt the 0.015 kg block of ice at 0ºC?
To melt the ice, we need to consider the specific heat for the phase change and the heat of fusion.
Q_melt = m2 * Lf
Where:
m2 = mass of ice (in kg)
Lf = heat of fusion (in kcal/kg), which is the amount of energy required to change the phase of 1 kg of a substance
Given:
m2 = 0.015 kg
Lf = 79.7 kcal/kg
Plugging in the values:
Q_melt = 0.015 kg * 79.7 kcal/kg = 1.1955 kcal
So, the heat required to melt the 0.015 kg block of ice at 0ºC is 1.1955 kcal.
(c) How much heat in kcal must be provided for a 0.853 kg block of lead to raise the temperature of 0.015 kg of liquid water from 0ºC to 5ºC?
To raise the temperature of water, we use the same specific heat formula as in part (a):
Q_water = m_water * c_water * ΔT_water
Where:
m_water = mass of water (in kg)
c_water = specific heat of water (in kcal/kg*Cº)
ΔT_water = change in temperature (in ºC)
Given:
m_water = 0.015 kg
c_water = 1 kcal/kg*Cº
ΔT_water = 5ºC - 0ºC = 5ºC
Plugging in the values:
Q_water = 0.015 kg * 1 kcal/kg*Cº * 5ºC = 0.075 kcal
So, the heat required to raise the temperature of 0.015 kg of liquid water from 0ºC to 5ºC is 0.075 kcal.
(d) What is the total amount of heat in kcal needed for the whole process?
To find the total heat required for the whole process, we sum up the results from parts (a), (b), and (c):
Total_heat = Q_total + Q_melt + Q_water
Given:
Q_total = 0.15675 kcal
Q_melt = 1.1955 kcal
Q_water = 0.075 kcal
Plugging in the values:
Total_heat = 0.15675 kcal + 1.1955 kcal + 0.075 kcal = 1.42725 kcal
So, the total amount of heat needed for the whole process is 1.42725 kcal.
(e) To what temperature must a 0.853 kg block of lead be heated to complete this task? (specific heat for lead = 0.0305 kg/kcal*Cº)
To find the final temperature, we can rearrange the specific heat formula:
Q_lead = m_lead * c_lead * ΔT_lead
Where:
Q_lead = heat absorbed by the block of lead (in kcal)
m_lead = mass of lead (in kg)
c_lead = specific heat of lead (in kcal/kg*Cº)
ΔT_lead = change in temperature of the lead (in ºC)
We want to find ΔT_lead when Q_lead is equal to the total heat required for the whole process from part (d).
Given:
Q_lead = 1.42725 kcal
m_lead = 0.853 kg
c_lead = 0.0305 kcal/kg*Cº
Rearranging the formula gives us:
ΔT_lead = Q_lead / (m_lead * c_lead)
Plugging in the values:
ΔT_lead = 1.42725 kcal / (0.853 kg * 0.0305 kcal/kg*Cº) = 57.3287 ºC
Therefore, the 0.853 kg block of lead must be heated to a temperature of 57.3287 ºC to complete this task.