Calorimetry....

Question

A 57.28 gram sample of iron (with a heat capacity of 0.450 ^J/_gºC) is heated to 100.0 ºC. It is then transferred to a coffee cup calorimeter containing 47.53 g of water (specific heat of 4.184 ^J/ _gºC) initially at 20.63 ºC. If the final temperature of the system is 23.59ºC, how much heat was absorbed by the bomb calorimeter? (Please give the absolute value)  

Answer 1

To determine the heat absorbed by the bomb calorimeter, we can use the equation:

q = m × c × ΔT

where:

q is the heat absorbed (in Joules),

m is the mass of the substance (in grams),

c is the specific heat capacity of the substance (in J/gºC), and

ΔT is the change in temperature (in ºC).

First, let's calculate the heat absorbed by the iron sample:

m_iron = 57.28 g

c_iron = 0.450 J/gºC

ΔT_iron = (100.0 ºC - 20.63 ºC) = 79.37 ºC

q_iron = m_iron × c_iron × ΔT_iron

= 57.28 g × 0.450 J/gºC × 79.37 ºC

= 2045.29 J

Next, we'll calculate the heat absorbed by the water in the calorimeter:

m_water = 47.53 g

c_water = 4.184 J/gºC

ΔT_water = (23.59 ºC - 20.63 ºC) = 2.96 ºC

q_water = m_water × c_water × ΔT_water

= 47.53 g × 4.184 J/gºC × 2.96 ºC

= 596.95 J

Since the heat absorbed by the bomb calorimeter is equal to the heat released by the iron sample (assuming no heat loss to the surroundings):

q_bomb_calorimeter = -q_iron = -2045.29 J (negative sign because heat is released)

Therefore, the heat absorbed by the bomb calorimeter is approximately 2045.29 Joules.