Question

When 36 g of a metal at 86 ◦C is added to

41 g of water at 30 ◦C, the temperature of the
water rises to 36 ◦C. What is the specific heat
capacity of the metal? Assume no heat was
lost to the surroundings.
Answer in units of J/g ·◦C
.

Answer 1

is there any answer choices

Answer 2

0.57J/g°C is the specific heat capacity of the metal.

Step-by-step explanation:

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

`-Q_1=Q_2`

Mass of metal=
`m_1=36 g`

Specific heat capacity of metal=
`c_1=?`

Initial temperature of the metal=
`T_1=86^oC`

Final temperature =
`T_2=T=36^oC`

`Q_1=m_1c_1* (T-T_1)`

Mass of water=
`m_2=41 g`

Specific heat capacity of water=
`c_2=4.18 J/g^oC`

Initial temperature of the water =
`T_3=30^oC`

Final temperature of water =
`T_2=36^oC`

`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:

`-(36 g* c_1* (36^oC-86^oC))=41 g* 4.18 J/g^oC* (36^oC-30^oC)`

we get,
`c_2 =0.57J/g^oC`

0.57J/g°C is the specific heat capacity of the metal.