An object with a mass of 50 g is dropped into 300 mL of water at 0^@C. If the object cools by 80 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
1s2s2p
Dec 7, 2017

C_o=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)

Explanation:

DeltaE=mcDeltatheta, where:

  • DeltaE = change in energy (J)
  • m = mass of the object (kg)
  • c = specific heat capacity of the object (J kg^(-1) K^(-1))
  • Deltatheta = change in temperature (K)

Assuming all heat is transferred we have:
m_oC_oDeltatheta_o=m_wc_wDeltatheta_w

m_o=50g=0.05kg
Deltatheta_o=80^circC=80K
rho_w~~1000kgcolor(white)(l)m^(-3)
V_w=300mL=0.3L=0.0003m^3
m_w=1000*0.0003=0.3kg
Deltatheta_w=8^circC=8K
c_w~~4180Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)

0.05C_o*80=0.3*4180*8

C_o=(0.3*4180*8)/(0.05*80)=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)

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