Question

An 8.42 g piece of metal with a specific heat of 1.020 J g-1 k-1 wasn't heated to an unknown temperature. The metal was then placed in 44.2 g of water with an initial temperature of 18.50 C. If the final temperature of the water was 20.40 C, what temperature was the metal initially heated to (in C).

Answer 1

Answer:

61.469 °C

Explanation:

From the question,

Heat lost by the metal = heat gained by water.

CM(t₁-t₃) = cm(t₃-t₂).................. Equation 1

Where C = Specific heat capacity of the metal, M = mass of metal, m = mass of water, c = specific heat capacity of water, t₁ = Initial temperature of metal, t₂ = initial temperature of water, t₃ = Final temperature of the mixture.

Make t₁ the subject of the equation

t₁ = [cm(t₃-t₂)/CM]+t₃................. Equation 2

Given: C = 1.020 J/g.K, M = 8.42 g, c = 4.2 J/g.K, m = 44.2 g, t₂ = 18.50 °C, t₃ = 20.40 °C

Substitute into equation 2

t₁ = [4.2×44.2(20.40-18.50)/(1.020×8.42]+20.40

t₁ = (352.716/8.5884)+20.40

t₁  = 41.069+20.40

t₁ = 61.469 °C

Answer 2

Answer:

$T = 61.344\,^{\textdegree}C$

Explanation:

The heat received by water is equal to the heat rejected by the piece of metal. That is to say:

$-Q_{piece} = Q_{water}$

$(8.42\,g) \cdot \left(1.020\,\frac{J}{g\cdot ^{\textdegree}C} \right) \cdot (T - 20.40\,^{\textdegree}C)= (44.2\,g)\cdot \left(4.187\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (20.40\,^{\textdegree}C-18.50\,^{\textdegree}C)$

$\left(8.588\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20.40\,^{\textdegree}C) = 351.624\,J$

The initial temperature of the piece of metal is:

$T = 61.344\,^{\textdegree}C$