A89.1 g piece of copper (specific heat 0.38 j/g･°c) is heated and then placed into 400.0 g of water initially at 20.7°c. the water increases in temperature to 22.2°c. what is the initial temperature of the copper? (the specific heat of water is 4.18 j/g･°c).

Here, we are required to determine the initial temperature of the copper.

The initial temperature of the copper is;

T(c1) = 96.27 °C

The energy required to raise the temperature of one gram of a material by one degree celsius is termed the Specific Heat Capacity of that material.

Mathematically, we have;

Q=mc{T(2) -T(1)}

Q=mc{T(2) -T(1)} where:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

m(c) × c(c) × (T2(c) - T1(c)) = -{m(w) × c(w) × (T2(w) - T1(w))}

However, the final temperature of the copper piece and water can be assumed to be equal

i.e. T2(c) = T1(w)

+89.1*0.38*(22.2-T(c1)) = -400*4.18(22.2-20.7)

(22.2−T(c1) ) = −2508/33.858

(22.2−T(c1) ) = -74.07

T(c1) = 74.07 + 22.2

T(c1) = 96.27 °C.

Therefore, the initial temperature of the copper is; T(c1) = 96.27 °C

Read more:

link;

i'm just gonna leave this comment here, so i can get a challange done-

explanation:

increase masses, decrease separation