Question

A block of copper was heated to 55oC and dropped into a beaker of 55 mL of water at 22oC. After the water reached thermal equilibrium, the temperature was 28oC. What was the mass of the copper block? Assume no heat is lost by the beaker.Cs, Cu= 0.385 J/gKCs, H2O= 4.18 J/gK

Answer 1

Answer:

132.70kg

Explanation:

Given:

Temperature of copper block = 55°C

Temperature of water = 22°C

Mass of copper block, m₁ = ?

Mass of water, m₂ = 55 mL = 5.5 × 10⁻⁵m³

                Density of water $= \frac{mass}{volume}$

                                   m₂  = Density of water × volume

                                   m₂ = 1000kg/m³ × (5.5 × 10⁻⁵m³)

                                   m₂ = 0.055 kg

                                    m₂ = 55g

Quantity of heat lost by the copper block = m₁cΔT

                                            = m₁ × 0.385 × (55 - 28)

                                            = m₁ × 0.385 × 27

                                            = 10.395 × m₁  J

Quantity of heat gained by the water = m₂cΔT

                                           = 55 × 4.18 × (28-22)

                                           = 55 × 4.18 × 6

                                           = 1379.4 J

Heat lost by copper block = Heat gained by water

                              10.395m₁ = 1379.4

                                $m_{1} =\frac{1379.4}{10.395}$  

                                   m₁ = 132.6984kg

                                   m₁ = 132.70kg

The mass of the copper block is 132.70kg