What is the specific heat of an object that weighs 25 g and when heated to a temp of 500K and dropped into 100mL of water changes the temp of the water from room temp (293K) to 306K? Cp of water= 4.184
heat lost by object + heat gained by water = 0
[mass object x specific heat object x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute the numbers from the problem and solve for specific heat object.
To calculate the specific heat of the object, we can use the equation:
q = m * c * ΔT
Where:
q is the heat absorbed or released by the object
m is the mass of the object
c is the specific heat capacity
ΔT is the change in temperature
In this case, the object is heated from 293K to 500K, resulting in a change in temperature of ΔT = 500K - 293K = 207K.
The heat absorbed by the water can be calculated using the formula:
q_water = m_water * Cp_water * ΔT_water
Where:
q_water is the heat absorbed by the water
m_water is the mass of the water
Cp_water is the specific heat capacity of water
ΔT_water is the change in temperature of the water
Since the water is heated from 293K to 306K, the change in temperature is ΔT_water = 306K - 293K = 13K.
Now, let's substitute the given values into the equations:
q = m * c * ΔT
q_water = m_water * Cp_water * ΔT_water
Given:
m = 25g (mass of the object)
ΔT = 207K (change in temperature of the object)
m_water = 100mL (volume of water) -> Since grams and milliliters have the same units, we can directly use 100g.
Cp_water = 4.184 (specific heat capacity of water)
ΔT_water = 13K (change in temperature of the water)
Solving the equation for the object and rearranging for c:
q = m * c * ΔT
c = q / (m * ΔT)
Substituting the values:
c = q / (m * ΔT)
c = q / (25g * 207K)
Now, we need to calculate the heat absorbed by the object, q.
Using the equation:
q = q_water
Since the heat lost by the object will be equal to the heat gained by the water for no energy loss.
q_water = m_water * Cp_water * ΔT_water
q = 100g * 4.184 * 13K
Calculating q:
q = 100g * 4.184 * 13K
q = 54592 cal
Substituting the value of q into the equation for c:
c = q / (25g * 207K)
c = 54592 cal / (25g * 207K)
Simplifying the equation:
c ≈ 10.57 cal/gK
Therefore, the specific heat of the object is approximately 10.57 cal/gK.