Question

A piece of metal with a mass of 1.4 kg, specific heat of 207 j/kg/ ◦c, and initial temperature of 97◦c is dropped into an insulated jar that contains liquid with a mass of 4 kg, specific heat of 1000 j/kg/ ◦c, and initial temperature of 0◦c. the piece of metal is removed after 6.7 s, at which time its temperature is 22◦c. find the temperature of the liquid after the metal is removed. neglect any effects of heat transfer to the air or to the insulated jar. answer in units of ◦c.

Answer 1

The temperature of the liquid after the metal is removed is 7.894 °C.

Step-by-step explanation:

The temperature of the liquid after the metal is removed can be calculated using the principle of energy conservation. We can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature. First, calculate the heat transferred by the piece of metal: Q1 = mcΔT = (1.4 kg)(207 J/(kg·°C))(22 °C - 97 °C) = -31575.6 J. Next, calculate the heat transferred by the liquid: Q2 = mcΔT = (4 kg)(1000 J/(kg·°C))(T - 0 °C), where T is the final temperature of the liquid. Since the inital temperature of the liquid is 0 °C, ΔT is T - 0 = T. Therefore, Q2 = 4000T. Since energy is conserved, Q1 + Q2 = 0, which gives -31575.6 J + 4000T = 0. Solving for T gives T = 7.8939 °C.