Question

An 8.4 g ice cube is placed into 260 g of water. Calculate the temperature change in the water upon the complete melting of the ice. Assume that all of the energy required to melt the ice comes from the water. Express your answer in terms of the initial temperature of water, T.a. -0.033T - 2.6 °Cb. -2.6T + 0.033 °Cc. 0.033T - 2.6 °Cd. 2.6T -0.033 °C

Answer 1

Answer:

c. 0.033T - 2.6 °C

Explanation:

Hello,

In this case, we should conclude that the energy lost by the water is gained by the ice to get melted, thus, we can write them in terms of melting and cooling enthalpies:

$\Delta H_{ice,melting}=-\Delta H_{water,cooling}$

First term includes melting enthalpy of ice that is 333.89 J/g and the second one specific heat of water that is 4.18 J/g°C, therefore, we obtain:

$8.4g*333.89\frac{J}{g}=-260g*4.18\frac{J}{g\°C}(T_2-T_1) \\\\(T_2-T_1) =-\frac{8.4g*333.89\frac{J}{g}}{260g*4.18\frac{J}{g\°C}} \\\\(T_2-T_1)=-2.6\°C$

Thus, answer should be c. 0.033T - 2.6 °C since it includes the temperature decrease of water due to the undergoing cooling.

Best regards.