Question

A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H2O(l) at 27.00°C in an insulated container. The temperature rises to 30.56°C. Neglecting the heat capacity of the container, what is the specific heat of the metal? The specific heat of H2O(l) is 4.18 J/(g ∙ °C). Group of answer choices 4.18 J/(g ∙ °C) 60.4 J/(g ∙ °C) 0.289 J/(g ∙ °C) 0.289 J/(g ∙ °C) 14.4 J/(g ∙ °C)

Answer 1

Answer:

The specific heat of the metal is 0.289 J/g°C

Explanation:

Step 1: Data given

Mass of metal = 110.0 grams

Temperature of the metal = 82.00 °C

MAss of water = 110.0 grams

Temperature of the water = 27.00 °C

The final temperature = 30.56 °C

The specific heat of H2O is 4.18 J/(g°C).

Step 2: Calculate the specific heat of the metal

Heat lost = heat gained

Qlost = - Qgained

Qmetal = - Qwater

Q =m*c*ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒with m(metal) = the mass of the metal = 110.0 grams

⇒with c(metal) = the specific heat of the metal = TO BE DETERMINED

⇒with ΔT(metal) = the change of temperature of the metal = T2 - T1 = 30.56 - 82.00 °C= -51.44 °C

⇒with m(water) = the mass of water = 110.0 grams

⇒with c(water) = the specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 30.56 °C - 27.00 °C = 3.56 °C

110.0 * c(metal) * -51.44 = -110.0 * 4.18 * 3.56

c(metal) = 0.289 J/g°C

The specific heat of the metal is 0.289 J/g°C

Answer 2

Answer:

$Cp_{metal}=0.289\frac{J}{g^oC}$

Explanation:

Hello,

In this case, as the water is cold (lower initial temperature) and the metal is hot (higher initial temperature), the heat lost by the metal is gained by the water to attain an equilibrium temperature of 30.56 °C, this in an equation turns out:

$-\Delta H_{metal}=\Delta H_{water}$

In such a way, in terms of masses, heat capacities and temperatures we have:

$-m_{metal}Cp_{metal}(T_{eq}-T_{metal})=m_{water}Cp_{water}(T_{eq}-T_{water})$

Hence, solving for the heat capacity of the metal:

$Cp_{metal}=\frac{m_{water}Cp_{water}(T_{eq}-T_{water})}{-m_{metal}Cp_{metal}(T_{eq}-T_{metal})}$

Thus, with the given data we obtain:

$Cp_{metal}=\frac{110.0g*4.18\frac{J}{g^oC} (30.56^oC-27.00^oC)}{-110.0g(30.56^oC-82.00^oC)}\\\\Cp_{metal}=0.289\frac{J}{g^oC}$

Best regards.