Question

A 65.06 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 32.49 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 calorimeter? (Please give the absolute value)

Answer 1

Answer:

1835 J  

Explanation:

There are three heat flows in this question.

Heat lost by Fe + heat gained by water + heat gained by calorimeter = 0

q₁ + q₂ + q₃ = 0

m₁C₁ΔT₁ + m₂C₂ΔT₂ + q₃  = 0

Data:

m₁ = 65.06 g

C₁ = 0.450 J·°C⁻¹g⁻¹

Ti = 100.0 °C

T_f = 23.59 °C

m₂ = 32.49 g

C₂ = 4.184 J·°C⁻¹g⁻¹

Ti = 20.63 °C

T_f = 23.59 °C

Calculations:

(a) Heat lost by iron

ΔT₁ = 23.59 °C - 100.0 °C = -76.41 °C

q₁ = m₁C₁ΔT₁ = 65.06 g × 0.450 J·°C⁻¹g⁻¹ × (-76.41 °C) = -2237 J

(b) Heat gained by water

ΔT₂ = 23.59 °C - 20.63 °C = 2.96 °C

q₂ = m₂C₂ΔT₂ = 32.49 g × 4.184 J·°C⁻¹g⁻¹ × 2.96  °C = 402 J

(c) Heat gained by calorimeter

-2237 J + 402 J + q₃ = 0

             -1835 J + q₃ = 0

                             q₃ = 1835 J

The heat gained by the calorimeter was 1835 J.