An object with a mass of $160 g$ is dropped into $880 mL$ of water at $0^@C$ . If the object cools by $120 ^@C$ and the water warms by $18 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-02-18T09:27:00

The specific heat is $= 3.453 KJkg^-1°C^-1$

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

For the cold water, $Delta T_w=18ºC$

For the object $DeltaT_o=120ºC$

$M_o C_o (DeltaT_o) = Mw C_w (Delta_w)$

$C_w=4.186KJkg^-1ºC^-1$

$0,160 C_o120 = 0,880* 4.186 *18$

$C_o = (0.880*4.186*18)/(0.160*120) = 3.453 KJkg^-1°C^-1$

An object with a mass of 160 g160 g is dropped into 880 mL880 mL of water at 0^@C0^@C. If the object cools by 120 ^@C120 ^@C and the water warms by 18 ^@C18 ^@C, what is the specific heat of the material that the object is made of?