Question

A 32.1-g ice cube at 0 °C is added to 120 g of water in a 64.2-g aluminum cup. The cup and the water have an initial temperature of 23.5 °C. Calculate the equilibrium temperature of the cup and its contents. Do not enter unit.

Answer 1

`T = 3.5^0 C`

Step-by-step explanation:

Heat given by water + cup = Heat absorbed by the ice

here we can say that let the final temperature of the system is "T"

so we will have heat absorbed by the ice given as

`Q = mL + ms\Delta T`

`Q = (32.1)(335) + (32.1)(4.186)(T - 0)`

`Q_(in) = 10753.5 + 134.4 T`

now we will have heat given by cup + water as

`Q = m_w s_w(23.5 - T) + m_c s_c(23.5 - T)`

`Q = 120(4.186)(23.5 - T) + 64.2(0.900)(23.5 - T)`

`Q_(out) = 560.1(23.5 - T)`

now we have

`Q_(in) = Q_(out)`

`10753.5 + 134.4T = 560.1(23.5 - T)`

`10753.5 + 134.4T = 13162.4 - 560.1 T`

`694.5T = 2409`

`T = 3.5^0 C`