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Marysya12 [62]
3 years ago
14

Determine the specific heat of a 213.0 g piece of metal in J/g•°C if the temperature of the metal drops from 70.2 °C to 25.6 °C

when placed in a calorimeter with 250.0 mL of water at an initial temperature of 22.5 °C (specific heat of water is 4.184 J/g•K).
Chemistry
1 answer:
ladessa [460]3 years ago
3 0

Answer:

0.3415 J/g °C

Explanation:

M1 = 213.0g

c1 = ?

T1 = 70.2°C

T3 = 25.6 °C

T2 = 22.5°C

C2 = 4.186 J/g°C

Vol. of water = 250mL = 250cm³

p (density of water ) = 1g/L

p = mass / volume

mass (m) = density (p) * volume (v)

m = 1 * 250 = 250g

Heat loss by metal = heat gain by water

Q1 = Q2

Mc∇T = Mc∇T

Mc (T1 - T3) = Mc* (T3 - T2)

213 * c * (70.2 - 25.6) = 250 * 4.186 * (25.6 - 22.5)

(213 * 44.6)c = 1046.5 * 3.1

9499c = 3244.15

c = 0.3415 J/g.°C

The specific heat capacity of the metal is 0.3415 J/g °C

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When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

150.0 mL \times \frac{1.02g}{mL}  = 153 g

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J  }{\frac{4.184J}{g.\° C }  \times 153g} = 5.87 \° C

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