Question

A 32-g chunk of chilled glass at 12°C is dropped into 54 g of water at 37°C. What is the equilibrium temperature of the system? The specific heats of glass and water are 0.840 and 4.186 J/(g×°C), respectively.

Answer 1

Answer:25c

Step-by-step explanation:

Answer 2

Answer: The equilibrium temperature of the system is 29.9°C

Step-by-step explanation:

When chilled glass is dropped in water, the amount of heat released by water will be equal to the amount of heat absorbed by glass.

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of glass = 32 g

`m_2` = mass of water = 54 g

`T_(final)` = final temperature = ? °C

`T_1` = initial temperature of glass = 12°C

`T_2` = initial temperature of water = 37°C

`c_1` = specific heat of glass = 0.840 J/g°C

`c_2` = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

`32* 0.840* (T_(final)-12)=-[54* 4.186* (T_(final)-32)]`

`T_(final)=29.9^oC`

Hence, the equilibrium temperature of the system is 29.9°C