# An object with a mass of 4 g is dropped into 650 mL of water at 0^@C. If the object cools by 300 ^@C and the water warms by 30 ^@C, what is the specific heat of the material that the object is made of?

Feb 18, 2017

The specific heat of the material is =68.0 KJkg^-1°C^-1

#### Explanation:

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

For the cold water,  Delta T_w=30ºC

For the object DeltaT_o=300ºC

${M}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = M w {C}_{w} \left({\Delta}_{w}\right)$

C_w=4.186KJkg^-1ºC^-1

$0 , 004 \cdot {C}_{o} \cdot 300 = 0 , 6500 \cdot 4.186 \cdot 30$

C_o = (0.650*4.186*30)/(0.004*300) = 68.0 KJkg^-1°C^-1