### Step 1 of 2

Given data: given is 60.80g of iron sample

heat capacity= 0.450 J/g ∘C

water= 52.42 g

specific heat = 4.184 J/ g◦C

Find: To find the heat gained by the calorimeter.

**Explanation:**

we can use the formula:

q calorimeter=m calorimeter×C calorimeter×ΔT calorimeter

### Step 2 of 2

q calorimeter=m calorimeter×C calorimeter×ΔT calorimeter

where:

*m*calorimeter is the mass of the calorimeter,*C*calorimeter is the specific heat capacity of the calorimeter,- Δ
*T*calorimeter is the change in temperature of the calorimeter.

*m *calorimeter=25.19g,

*C *calorimeter is not given directly, but we can assume it’s the same as water (4.184J/g °C), and

Δ*T*calorimeter can be calculated as *T *final−*T *initial.

Δ*T*calorimeter=*T *final−*T *initial

The final temperature (*T *final) is 28.78°*C* and the initial temperature (*T *initial) is 20.47°*C*,

hence, we can calculate Δ*T*calorimeter.

After finding *q *calorimeter, we can then calculate the heat capacity of the calorimeter using the formula:

C calorimeter= q calorimeterΔTcalorimeter

Now, let’s calculate these values:

Δ*T*calorimeter= 28.78°C−20.47°C

*q *calorimeter= 25.19×4.184×(28.78−20.47)

C calorimeter= q calorimeterΔT calorimeter

C calorimeter = 25.19×4.184×(28.78−20.47)28.78°C−20.47°C

C calorimeter=0.6019 Jg∘C

**Explanation:**

Performing the calculations will give you the values for *q*calorimeter and *C*calorimeter.

##### Final solution

Therefore, the heat gained by the calorimeter is = 0,6019 J/g^{o}C