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?

a. qm = 627.3 J

b. qw = 627.3 J

c. C₂ = 227.4 J/kg.°C

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

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !