Question

A volume of 75.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C)

Answer 1

Answer: 17.78g

Explanation:

Assume there is no heat exchange with the environment, then the amount of heat taken by the steel rod, Q(s), is equal to the amount of heat lost by the water, Q(w), but with opposite sign.

Q(s) = -Q(w)

Remember, Q = mc(ΔΦ)

Where Q = amount of heat

m = mass of steel

c = specific heat capacity of steel

ΔΦ = Initial temperature T1 - Final temperature T2

Q = mc(T1-T2)

Recall, Q(s) = -Q(w). Then,

m(s)*c(s)*(T1s - T2s) = - m(w)*c(w)*(T1w - T2w)

Substituting each values

Note: m(w) = volume of water*density = 75mL*1g/mL = 75g

m(s)*0.452*(21.5-2) = -75*4.18*(21.5-22)

m(s)*8.814 = 156.75

m(s) = 156.75/8.814

m(s) = 17.78g

Therefore, the mass of steel is 17.78g

Answer 2

Specific heat capacity is the quantity of heat that is needed to raise the temperature of 1 kg of substance by $\bold{ 1^oC}$.  The mass of the steel rod is 17.78 g.

The Specific heat capacity:

It the amount of heat needed to increase the temperature per unit mass. It is given by the formula ,

$\bold{Q = mc \Delta T}$

where,

Q - energy absorbed

m - mass

$\bold{\Delta T}$ - temperature difference

Given here,

The heat released by the rod is equal to the heat absorbed by water.

Qr= -Qw

Put the values in the heat capacity formula,

$\bold {m(r)\times0.452\times (21.5-2) = -75\times 4.18\times (21.5-22)}\\\\\bold {m(r)\times 8.814 = 156.75 }\\\\\bold {m(r) = \dfrac{ 156.75}{8.814} }\\\\\bold {m(r) = 17.78g}$

Therefore, the mass of the steel rod is 17.78 g.

To know more about specific heat capacity, refer to the link:

https://brainly.com/question/11194034