A 0.2 kg piece of unknown metal is heated to 87 °C and then dropped into a 0.28 kg sample of water at 22°C. If the final temperature of the mixture is 24°C, what is the specific heat of the unknown metal?

1 Answer
Mar 2, 2017

Answer:

The specific heat of the metal is #=0.186kJkg^-1K^-1#

Explanation:

The heat transferred from the hot metal, is equal to the heat absorbed by the cold water.

For the cold water, # Delta T_w=24-22=2º#

For the metal #DeltaT_o=87-24=63º#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186kJkg^-1K^-1#

#m_0 C_o*63 = m_w* 4.186 *2#

#0.2*C_o*63=0.28*4.186*2#

#C_o=(0.28*4.186*2)/(0.2*63)#

#=0.186kJkg^-1K^-1#