Question

A 13.5 g sample of gold is heated to 125.0°C, then placed in a calorimeter containing 60.0 g of water. The final temperature of the water 20.00 o C. The specific heat of gold is 0.130 J/g o C. What was the initial temperature of the water?

Answer 1

The initial temperature was 19.27 °C.

The guiding principle is the Law of Conservation of Energy: the sum of all the energy transfers must add up to zero.

The formula for the heat q gained or lost by a substance is

q = mCΔT

where

m = the mass of the substance.

C = its specific heat capacity.

ΔT = T_f - T_i = the change in temperature.

In this problem, there are two heat transfers.

Heat lost by gold + heat gained by water = 0

m _1C_1ΔT_1 + m_2c_2ΔT_2= 0

m_1 = 13.5 g; C_1 = 0.130 J·°C^(-1)g^(-1); ΔT_1 = T_f – T_i = 20.00 °C – 125.0 °C = -105.0 °C

m_2 = 60.0 g; C_2 = 4.184 J·°C^(-1)g^(-1); ΔT_2 = ?

q_1 = m_1C_1ΔT_1 = 13.5 g × 0.130 J·°C^(-1)g^(-1) × -105.0 °C = -184.3 J

q_2 = m_2C_2ΔT_2 = 60.0 g × 4.184 J·°C^(-1)g^(-1) × ΔT_2

= 251.0 ΔT_2 J·°C^(-1)

q_1+ q_2 = -184.3 J + 251.0 ΔT_2 J·°C^(-1) = 0

251.0 ΔT_2 °C^(-1) = 184.3

ΔT_2 = 184.3/251.0 °C^(-1) = 0.734°C

ΔT_2 = T_f - T_i = 20.00 °C -T_i = 0.734 °C

T_i = 20.00 °C – 0.734 °C = 19.27 °C