Question

If 200 cm^3 of tea at 95 degrees celsius is poured into a 150g glass cup initially at 25 degrees celsius, what will be the common final temperature T of the tea and cup when equilibrium is reached, assuming no heat flows to the surroundings?

Answer 1

Answer:

The final temperature is 85.86°C.

Explanation:

Given that,

Mass of tea = 200 g

Mass of glass = 150 g

Initial temperature = 25°C

We need to calculate the final temperature

When equilibrium is reached, assuming no heat flows to the surroundings

So,

$Q_{glass}+Q_{tea}=0$

$m_{g}C\Delta T+m_{g}C\Delta T=0$

We know that,

Specific heat of tea = 1.00 cal/gm° C

Specific heat of tea = 0.20 cal/gm° C

Put the value into the formula

$150\times0.20\times(T_{2}-T_{1})+200\times1.00\times(T_{2}-T_{1})$

$30T_{2}-30T_{1}+200T_{2}-200T_{1}=0$

$30T_{2}+200T_{2}=30T_{1}+200T_{1}$

Put the value of initial temperature

$T_{2}=\dfrac{30\times25+200\times95}{230}$

$T_{2}=85.86^{\circ}C$

Hence, The final temperature is 85.86°C.