An object with a mass of #30 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #20 ^@C# and the water warms by #80 ^@C#, what is the specific heat of the material that the object is made of?

Question

2031 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-20T17:56:52

#Delta T_w=80°C#
#Delta T=20°C#
#m=30 g#
#V=750 m L#

#Q_(water)=Q_(object)#
#Q=m c_p Delta T#
#m_w c_p^w Delta T_w=m c_p Delta T#
#c_p=m_w/m (Delta T_w)/(Delta T) c_p^w#

#rho_w=1000 (kg)/m^3#
#c_p^w=4150 J/(kg K)#
#m_w=V* rho_w=750*10^(-3)*10^(-3)m^3*1000 (kg)/m^3=0.75 kg#

#c_p=m_w/m (Delta T_w)/(Delta T) c_p^w=#
#=(0.75 kg)/(0.03 kg) (80+273 K)/(20+273 K) 4150 J/(kg K)=125 (k J)/(kg K)#