A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water and calorimeter were initially at 21.0oC. If the final temperature of mixture was 30.5oC, calculate the specific heat capacity of the metal. The specific heat of water is 4.184 J/(g.oC) and heat capacity of calorimeter is 10.0 J/oC.

Because here temperature of metal is high that is why it loose the heat.The temperature of water and calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

Q= m Cp ΔT

m=Mass

Cp=Specific heat capacity

ΔT=Temperature difference

By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

Cp ( 99 .5- 30.5) = 30.65

Cp= 0.44 J/g.C

This is heat capacity of metal.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-29T00:43:22+02:00

The specific heat capacity of the hot metal sample is 0.444 J/g⁰C.

The given parameters;

mass of the metal = 55 g
initial temperature of the metal = 99.5 ⁰C
mass of water = 40 g
initial temperature of the water = 21 ⁰C
final temperature of the mixture = 30.5 ⁰C
specific heat capacity of the water = 4.184 J/g⁰C
specific heat capacity of calorimeter = 10.0 J/⁰C

Apply the principle of conservation of energy to determine the specific heat capacity of the metal.

Heat lost by the metal = Heat gained by the water and calorimeter

[tex]m_mc_m \Delta\theta_m= m_wc_w \Delta\theta+ C_{cal}\Delta\theta\\\\m_mc_m \Delta\theta_m = (m_wc_w + C_{cal})\Delta\theta\\\\55\times c_m \times (99.5- 30.5) = (40\times 4.184 \ \ + \ \ 10)(30.5-21)\\\\3795c_m = 1684.92\\\\c_m= \frac{1684.92}{3795} \\\\c_m = 0.444 \ J/g^o C[/tex]

Thus, the specific heat capacity of the hot metal sample is 0.444 J/g⁰C.

Learn more here:https://brainly.com/question/20709115