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
1.11 Calorie
Explanation:
Given,volume of water
Suppose,the specific heat of the object be
Solving we get,
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)
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