Chemistry, 24.07.2019 06:00 yangyang7

Apiece of metal of mass 12 g at 107 ◦c is placed in a calorimeter containing 55.7 g of water at 20◦c. the final temperature of the mixture is 27.6 ◦c. what is the specific heat capacity of the metal? assume that there is no energy lost to the surroundings.

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When the heat lost by metal q1 = the heat gained by water q2

and when the heat energy is Q = M*C*ΔT

∴ Mm*Cm*ΔTm = Mw*Cw*ΔTw

when Mm is the mass of metal sample = 12 g

Cm is the specific heat capacity of the metal

ΔTm the difference in temperature in metal = 107-27.6 = 79.4 °C

and:

Mw is the mass of water sample = 55.7 g

Cw is the specific heat capacity of water = 4.18 J.g/°C

ΔTw is the difference in temperature in water = 27.6 - 20 = 7.6 °C

so by substitution, we will get the Cm:

12g * Cm * 79.6 °C = 55.7 g* 4.18 J.g/°C * 7.6°C

∴Cm = (55.7*4.18*7.6) / (12*79.6)

= 1.85 J.g/°C

∴ the specific heat capacity of the metal = 1.85 J.g/°C

Answer from: cmcdonnell6945

the specific heat capacity of the metal : 1,866 J / kg C Further explanation

The law of conservation of energy can be applied to heat changes, ie the heat received / absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

A calorimeter is a device used to measure the specific heat of material

A metal is put into a calorimeter that contains water and there will be heat transfer:

$\displaystyle m_mc_m (T_m-T)=m_wc_w(T-Tw)$

m = metal

w = water

T = the final temperature of the mixture

known :

m of metal = 12 g = 0.012 kg

Tm = 107 ◦C + 273 = 380 K

m of water = 55.7 g = 0.0557 kg

Tw = 20 ◦C + 273 = 293 K

cw= 4,200 J / kg C

T = 27.6 + 273 = 300.6 K

the specific heat capacity of the metal :

$\displaystyle m_mc_m (T_m-T)=m_wc_w(T-Tw)$

$\displaystyle 0.012.c_m (380-300.6)=0.0557.4200(300.6-293)$

$\displaystyle c_m=1.866\:J/kg.C$

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Keywords: heat, temperature,calorimeter

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Apiece of metal of mass 12 g at 107 ◦c is placed in a calorimeter containing 55.7 g of water at 20◦c...

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