Answer to Question #156750 in General Chemistry for rome

Question

1. A 15.5 g piece of metal, heated to 100.0 ºC, is dropped into a calorimeter containing 55.5 g of water at 16.5 ºC. The final temperature of the metal and the water is 18.9 ºC. What is the specific heat capacity of the metal (Cs metal)? As part of your answer, complete the rest of the table with values for water on the left and the metal values on the right.

Water Metal

q =                 q = 

Cs =             Cs =

m =          m =

Tfinal =

Tfinal =  

Tinitial =  

Tinitial =  

ΔT =        ΔT =  

Can you also help understand this.

Answer 1

mass of the metal is 15.5g (m_m = 15.5g)

T₁ = 100°C

mass of water is 55.5g ( m_w = 15.5g)

T₂ = 16.5°C

final temperature is 18.9°C (T₃ = 18.9°C)

heat capacity of the metal is unknown (c_s = ?)

Change of temperature of metal (∆T_m = T₁ - T₃ = 100°C - 18.9°C = 81.1°C)

Change of temperature of water (∆T_w = T₃ - T₂ = 18.9°C - 16.5°C = 2.4°C)

specific heat capacity of water (c) = 4.18Jg^-1K^-1

Heat lost by metal = Heat gained by water$q_m = q_w$

$m_mc_s ∆T_m = m_wc∆T_w$

15.5 × c_s × 81.1 = 55.5 × 4.18 × 2.4

c_s = $\dfrac{55.5×4.18×2.4}{15.5×81.1}$ = 0.45Jg^-1K^-1