An object with a mass of $40 g$ is dropped into $350 mL$ of water. If the object cools by $25 ^@C$ and the water warms by $10 ^@C$ , what is the specific heat of the material that the object is made of?

Question

1997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-26T12:29:39

$3.5$ Cal per gram per degrees Celsius

$m_(water)*C_(water)*DeltaT=m_(object)*C_(object)*DeltaT$

$350*1*10 = 40*C_(object)*25$

$C_(object) = 3500/(40*25)$

$C_(object) = 3.5$ $(Cal)/(g*degrees C)$

Your answer:

$C_(object) = 3.5$ $(Cal)/(g*degrees C)$

An object with a mass of 40 g40 g is dropped into 350 mL350 mL of water. If the object cools by 25 ^@C25 ^@C and the water warms by 10 ^@C10 ^@C, what is the specific heat of the material that the object is made of?