Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.260kg and at temperature of 60.1C .
How much ice at -18.1 C must be dropped into the water so that the final temperature of the system will be 40.0C?
Take specific heat of liquid water to be 4190J/kg.K
the specific heat of ice to be 2100 J/kg.K,
and the heat of fusion for water to be 3.34 x10^5 J/kg
Please, help

Answer 1

\[M_w C_w \triangle T_w + M_i C_i \triangle T_i + M_i L_f=0\]
where \(M_w \) is mass of water = 0.260kg
\(C_w\) = 4190 J/kg.K
\(\triangle T_w= -20.1^0 C\)
\(\triangle T_i = 58.1^0 C\)
I have equation:
\[M_i = \frac{-M_w C_w\triangle T_w}{C_i\triangle T_i + L_f} = 0.048 kg\]
But they said It's wrong. I don't know how to solve.

Answer 2

@douglaswinslowcooper

Answer 3

@ybarrap

Answer 4

second time around, I tried:
\[Q_w = Q_1 + Q_2\]where
\[ Q_1 = M_w C_w \triangle T_w = 0.260*4190*(-20.1) = -21896\]
\[ Q_2 = -M_w L_f = -0.260*3.33*10^5=-86580\]
\[Q_w = -21896-86580=-108476.94\]
Now, \(Q_{ice}= M_{ice}C_{ice}\triangle T_{ice} \)
set the equation to get \(M_{ice} = \dfrac{108476.94}{2100*58.1}= 0.889kg\)
I am still wrong. haaaaaaa!!

Answer 5

is it aprox 40.58 g?

Answer 6

Heat lost by water = heat gained by ice

heat lost by water = 0.260 * 4190 * (60.1-40) = 21896.94 - 21897 J or 21.9 kJ

Now heat gained by ice is calculate in three steps
first -18 degree ice temp raises to 0 degree ice
then ice gets converted to water at 0 degree
then water at 0 degree gets heated to water at 40 degree.