Question

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final temperature of 51.9°C. The specific heat capacity of water is 4.186 joules/gram degree Celsius, and the specific heat capacity of limestone is 0.921 joules/gram degree Celsius. What was the initial temperature of the limestone? Express your answer to three significant figures.

The initial temperature of the limestone was

Answer 1

208.7°C was the initial temperature of the limestone.

Step-by-step explanation:

Answer 2

208.7°C was the initial temperature of the limestone.

Step-by-step explanation:

Heat lost by limestone will be equal to heat gained by the water

`-Q_1=Q_2`

Mass of limestone =
`m_1=62.6 g`

Specific heat capacity of limestone =
`c_1=0.921 J/g^oC`

Initial temperature of the limestone =
`T_1=?`

Final temperature =
`T_2`=T = 51.9°C

`Q_1=m_1c_1* (T-T_1)`

Mass of water=
`m_2=75.0 g`

Specific heat capacity of water=
`c_2=4.186 J/g^oC`

Initial temperature of the water =
`T_3=23.1^oC`

Final temperature of water =
`T_2`=T = 51.9°C

`Q_2=m_2c_2* (T-T_3)`

`-Q_1=Q_2`

`-(m_1c_1* (T-T_1))=m_2c_2* (T-T_3)`

On substituting all values:

`-(62.6 g* 0.921 J/g^oC* (51.9^oC-T_1))=75.0 g* 4.186 J/g^oC* (51.9^oC-23.1^oC)`

`T_1=208.7^oC`

208.7°C was the initial temperature of the limestone.