Question

You heat a 5.05 g piece of titanium to 98.2 oC and place it into 20.00 mL of room temperature water (24.8 oC ). The temperature of the water rises to 27.3 oC. The specific heat of water is 4.184 J/goC. The density of water is 0.997 g/mL. A. How much heat is absorbed by the water (in units of J)

Answer 1

Answer: The heat absorbed by the water is 52.823 J.

Step-by-step explanation:

Given: Mass of metal = 5.05 g

Specific heat of water = 4.184
`J/g^(o)C`

Initial temperature =
`24.8^(o)C`

Final temperature =
`27.3^(o)C`

Formula used to calculate heat absorbed is as follows.

`q = m * C * (T_(2) - T_(1))`

where,

q = heat

m = mass of substance

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\= 5.05 g * 4.184 J/g^(o)C * (27.3 - 24.8)^(o)C\\= 52.823 J`

Thus, we can conclude that heat absorbed by the water is 52.823 J.