Question

In the laboratory, a student determines the specific heat of a metal. She heats 18.6 grams of titanium to and then drops it into an insulated cup containing 80.5 grams of water at . When thermal equilibrium is reached, she measures the final temperature to be . Assuming that all of the heat is transferred to the water, she calculates the specific heat of titanium to be _____

Answer 1

Answer: 0.522 J/g°C

Step-by-step explanation:

To solve for the specific heat of titanium, we can use the formula:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

First, we need to calculate the amount of heat that is transferred from the titanium to the water:

Q = mTi * cTi * ΔTTi + mH2O * cH2O * ΔTH2O

Where mTi is the mass of titanium, cTi is the specific heat of titanium, mH2O is the mass of water, cH2O is the specific heat of water, ΔTTi is the change in temperature of the titanium, and ΔTH2O is the change in temperature of the water.

Since the experiment is carried out in an insulated cup, we can assume that there is no heat loss to the surroundings. Therefore, the amount of heat lost by the titanium is equal to the amount of heat gained by the water:

mTi * cTi * ΔTTi = - mH2O * cH2O * ΔTH2O

Note that we have placed a negative sign in front of the water's heat capacity because it is gaining heat, whereas the titanium is losing heat.

Now we can substitute the values given in the problem:

mTi = 18.6 g
cH2O = 4.18 J/g°C
mH2O = 80.5 g
ΔTH2O = 68.0°C - 22.4°C = 45.6°C

To find ΔTTi, we need to use the fact that the final temperature of the system is 30.5°C:

mTi * cTi * (30.5°C - 100°C) = - mH2O * cH2O * (30.5°C - 22.4°C)

Solving for cTi, we get:

cTi = [mH2O * cH2O * (30.5°C - 22.4°C)] / [mTi * (100°C - 30.5°C)]

cTi = [80.5 g * 4.18 J/g°C * 8.1°C] / [18.6 g * 69.5°C]

cTi = 0.522 J/g°C

Therefore, the specific heat of titanium is approximately 0.522 J/g°C.