Question

A 0.1000 kg brass block at 100.0°C is placed in 0.2000 kg of water at

20.0°C. The specific heat of brass is 376 J/kg•°C, the specific heat
capacity of liquid water is 4180 J/kg-°C. Assuming no heat loss to the
surroundings, what is the final temperature of the mixture? *

Pls help I’ll do anything

Answer 1

Tfinal = 23.4 [°C]

Step-by-step explanation:

In order to solve this problem we must perform an energy balance where the thermal energy of the brass block is transferred to the water, in this way it raises the water temperature to a value that we must calculate.

`Q_(B)=Q_(w)`

where:

QB = heat disipated by the block [J]

Qw = heat received by the water [J]

The thermal energy equation is defined as:

`Q=m*C_(p)*(T_(f)-T_(o))`

where:

m = mass of the element [kg]

Cp = specific heat of the element [J/kg*°C]

Tfinal = final temperature [°C]

To = initial temperature [°C]

`0.1*376*(100-T_(f))=0.2*4180*(T_(f)-20)\\3760-37.6*T_(f)=836*T_(f)-16720\\3760+16720=836*T_(f)+37.6*T_(f)\\20480=873.6*T_(f)\\T_(f)=23.44[C]`

The key to understanding the energy balance is that the final temperature of the block will be equal to the final temperature of the water so that there is thermal equilibrium. In this way we can clear the temperature value.