Question

An ice cute at 0.0*C was dropped into 30.0 g of water in a

cupat 45.0*C. At the instant that all of the ice was melted,
thetemperature of the water in the cup was 19.5*C. What was themass
of the ice cube?

Answer 1

Answer: The mass of ice cube is 77.90 grams

Step-by-step explanation:

When ice is mixed with water, the amount of heat released by ice will be equal to the amount of heat absorbed by water.

`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,

`m_1` = mass of ice = ? g

`m_2` = mass of water = 30.0 g

`T_(final)` = final temperature = 19.5°C

`T_1` = initial temperature of ice = 0.0°C

`T_2` = initial temperature of water = 45.0°C

`c_1` = specific heat of ice = 2.108 J/g°C

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

Putting values in equation 1, we get:

`m_1* 2.108* (19.5-0)=-[30.0* 4.186* (19.5-45.0)]`

`m_1=77.90g`

Hence, the mass of ice cube is 77.90 grams