An object with a mass of $150 g$ is dropped into $850 mL$ of water at $0^@C$ . If the object cools by $40 ^@C$ and the water warms by $3 ^@C$ , what is the specific heat of the material that the object is made of?

Question

1544 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-03T09:58:08

The specific heat of the object is $=1.78kJkg^-1K^-1$

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

For the cold water, $Delta T_w=3º$

For the metal $DeltaT_o=40º$

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

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

$m_0 C_o*40 = m_w* 4.186 *3$

$0.15*C_o*40=0.85*4.186*3$

$C_o=(0.85*4.186*3)/(0.15*40)$

$=1.78kJkg^-1K^-1$

An object with a mass of 150 g150 g is dropped into 850 mL850 mL of water at 0^@C0^@C. If the object cools by 40 ^@C40 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?