Question

A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal ( s = 0.900 J/g·°C), find the final temperature of the block and the water

Answer 1

34.17°C

Step-by-step explanation:

Given:

mass of metal block = 125 g

initial temperature
`T_i` = 93.2°C

We know

`Q = m c \Delta T` ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in
`J/g^(\circ)C`

`\Delta T` = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal =
`125* 0.9* (93.2-T_f)`

Heat gained by the water =
`100* 4.184*(T_f -18.3)`

thus, we have

`125* 0.9* (93.2-T_f)` =
`100* 4.184*(T_f -18.3)`

`10485-112.5T_f = 418.4T_f - 7656.72`

⇒
`T_f = 34.17^oC`

Therefore, the final temperature will be = 34.17°C