Question

A 10.0 g piece of hot metal at 300. °C was dropped into a 150.0 g sample of cooler room temperature water that was initially 25.0 °C. If 6.25 kJ of heat was transferred, what was the final temperature of the water?What was the specific heat of the metal?

Answer 1

Answer:

The final temperature of water is 34.92 ⁰C

The specific heat of the metal is 2357.78 J/kg⁰C

Explanation:

Given;

mass of the hot metal, m = 10.0 g = 0.01 kg

temperature of the hot metal, $t_m$ = 300 °C

mass of water, $m_w$ = 150 g = 0.15 kg

initial temperature of the water, $t_w_i$ = 25.0 °C

heat lost by the hot metal = heat gained by water

Q = 6.25 kJ = 6250 J

let the final temperature of water = T

$Q = m_w C_p_w (T-t_w_i)\\\\T-t_w_i = \frac{Q}{m_wC_p_w}\\\\ T= \frac{Q}{m_wC_p_w} + t_w_i\\\\T = \frac{6250}{(0.15)(4200)} + 25\\\\T = 34.92 ^0 C$

The final temperature of water is also the equilibrium temperature.

The specific heat of the metal is given by

$Q = mC_p_m (300 - T)\\\\C_p_m = \frac{Q}{m(300 - T)}\\\\ C_p_m = \frac{6250}{0.01(300 - 34.92)}\\\\ C_p_m = 2357.78 \ J/kg ^0C$