# An object with a mass of #120 g# is dropped into #800 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #5 ^@C#, what is the specific heat of the material that the object is made of?

##### 3 Answers

#### Answer:

1.11 Calorie

#### Explanation:

Given,volume of water

Suppose,the specific heat of the object be

Solving we get,

#### Answer:

#### Explanation:

The important concept here is that

Since the mass of the water is not given, we can derive it through its volume at a given temperature as provided. Knowing these, mass of the water can be calculated using the density formula; i.e.,

#Density(rho)=("mass"(m))/("Volume"(V))#

#m=rhoxxV# where:

#V=800ml " at " 0^oC# https://hypertextbook.com/facts/2007/AllenMa.shtml

#m=(0.9998g)/cancel(ml)xx800cancel(ml)#

#m=799.84g#

When the object was submerged in the water, temperature of the water has increased to 5^o that signifies heat

#Q=mCpDeltaT# where:

#m_(H_20)=799.84g#

#T_i=0^oC#

#T_f=5^oC#

#Cp_w=(4.18J)/(g*^oC)#

#Q=799.84cancel(g)xx(4.18J)/cancel((g*oC))xx5cancel(*^oC)#

#Q=16,716.66J#

Remember that

#Q=mCpDeltaT#

#Cp=(Q)/(mDeltaT)# where:

#m=120g#

#DeltaT=30^oC#

#Cp=(16,716.66J)/(120gxx30^oC#

#Cp=(4.64J)/(g*^oC)#

#### Answer:

The specific heat is

#### Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water,

For the object

The specific heat of water

Let,

The mass of the object is

The mass of the water is