Question

When 38.1 grams of a certain metal at a temperature of 90.0°C is added to 100.0mL of water at a temperature of 17.6°C in a perfectly insulated calorimeter, the final temperature of the resulting mixture is 19.1°C. a. Find the value of q for the metal b. Find the value of q for the water c. What is the specific heat of the metal?

Answer 1

a. qm = 627.3 J

b. qw = 627.3 J

c. C₂ = 227.4 J/kg.°C

Step-by-step explanation:

a.

Since, the calorimeter is completely insulated. Therefore,

Heat Lost by Metal = Heat Gained by water

qm = qw

qm = m₁C₁ΔT₁

where,

qm = heat lost by metal = ?

m₁ = mass of water = (density)(volume) = (1000 kg/m³)(100 mL)(10⁻⁶ m³/1 mL)

m₁ = 0.1 kg

C₁ = specific heat capacity of water = 4182 J/kg.°C

ΔT₁ = Change in Temperature of Water = 19.1°C - 17.6°C = 1.5°C

Therefore,

qm = (0.1 kg)(4182 J/kg.°C)(1.5°C)

qm = 627.3 J

b.

Since,

qm = qw

qw = 627.3 J

c.

qm = m₂C₂ΔT₂

where,

m₂ = mass of metal = 38.1 g = 0.0381 kg

C₂ = specific heat capacity of metal = ?

ΔT₂ = Change in Temperature of metal = 90°C - 17.6°C = 72.4°C

Therefore,

627.3 J = (0.0381 kg)(C₂)(72.4°C)

(627.3 J)/(0.0381 kg)(72.4°C) = C₂

C₂ = 227.4 J/kg.°C