Question

A piece of copper of mass 150g at temperature of mass 70g containing 60g of water at temperature of 20°C. Ignoring heat losses, what would be the final steady temperature of the mixture? [Specific of heat capacity of copper = 4.0× 10²J/kg-1/K ^-1. Please any Physics student can help or if you have any idea, you can help me out. Thanks.​✨✨​

Answer 1

The final temperature is
`29.6^oC`.

Step-by-step explanation:

When the two masses come in contact, one releases heat and the other absorbs it. The former can be modelled with the equation
`HeatLost = (Mass1 (kg))(c1)(T1-T)`, and the latter by
`HeatGained=(Mass2(kg))(c2)(T-T2)`

m1=0.15 kg

m2=0.06 kg

T1 = 70 degrees Celsius = 343 K

T2 = 20 degrees Celsius = 293 K

T= Final temperature

c1 = Specific heat capacity of copper

c2 = Specific heat capacity of water

Because there is no heat lost into the surroundings, the heat removed from one substance is the same as the heat gained in the other. Therefore:

`(Mass1)(c1)(T1-T)=(Mass2)(c2)(T-T2)`

`(0.150)(400)(343-T)=(0.06)(4184)(T-293)`

`(60)(343-T)=(251.04)(T-293)`

`20580-60T=251.04T-73554.72`

`-311.04T=-94134.72`

`T=302.6 K`

`T=29.6^oC`

Hope this helps! (My apologies if the answer is wrong, it has been a while since I've done this)