When you combine two different substances that may or may not have the same mass, you must use the concepts of Conservation of Energy to solve for how much the temperature of each substances will change. When discussing combining two substances, assuming no heat is transferred to the surroundings, you can assume that the heat lost by one substance is going to equal the heat gained by the other.
Conservation of Energy
\(\mathsf{ Q_{lost} + Q_{gained} = 0 }\)
\(\mathsf{ Q_{lost} = -Q_{gained} }\)
\(\mathsf{ (mc \Delta T)_1 = -(mc \Delta T)_2 }\)
You will need to know the specific heat capacity of the substances that are involved as well. Here are a few to reference as you work through the problems below.
| Substance | Specific Heat Capacity (J/kg °C) |
|---|---|
| water (liquid) | 4186 |
| aluminum | 899 |
| lead | 129 |
| iron | 412 |
| silver | 239 |
| gold | 129 |
Try these practice problems.
Silver + Water
Lead + Water
Gold + Water
What is the final temperature when a 3.2 kg silver bar at 97°C is dropped into 0.34 kg of water at 28°C?
\(\mathsf{ (mc \Delta T)_1 = -(mc \Delta T)_2 }\)
\(\mathsf{ (3.2 \text{ kg})(239 \text{ J/kg °C})(T_f - 97) = -(0.34 \text{ kg})(4186 \text{ J/kg °C})(T_f - 28) }\)
\(\mathsf{ (764.8)(T_f - 97) = -(1423.24)(T_f - 28) }\)
\(\mathsf{ 764.8T_f - 74185.6 = -1423.24T_f + 39850.72 }\)
\(\mathsf{ 764.8T_f + 1423.24T_f = 39850.72 + 74185.6 }\)
\(\mathsf{ 2188.04T_f = 114035.72 }\)
\(\mathsf{ T_f = \frac{114035.72}{2188.04} }\)
\(\mathsf{ T_f = 52 \text{°C} }\)
A 0.340 kg piece of lead initially at 0.00°C is dropped into 0.450 kg of water initially at 22.0°C. What is the final temperature?
\(\mathsf{ (mc \Delta T)_1 = -(mc \Delta T)_2 }\)
\(\mathsf{ (0.340 \text{ kg})(129 \text{ J/kg °C})(T_f - 0.00) = -(0.405 \text{ kg})(4186 \text{ J/kg °C})(T_f - 22.0) }\)
\(\mathsf{ (43.86)(T_f - 0.00) = -(1883.7)(T_f - 22.0) }\)
\(\mathsf{ 43.86T_f = -1883.7T_f + 41441.4 }\)
\(\mathsf{ 43.86T_f + 1883.7T_f = 41441.4 }\)
\(\mathsf{ 1927.56T_f = 41441.4 }\)
\(\mathsf{ T_f = \frac{41441.4}{1927.56} }\)
\(\mathsf{ T_f = 21.5 \text{°C} }\)
What is the final temperature when a 1.59 kg piece of gold at 35.0°C is dropped into 0.750 kg of water at 12.0°C?
\(\mathsf{ (mc \Delta T)_1 = -(mc \Delta T)_2 }\)
\(\mathsf{ (1.59 \text{ kg})(129 \text{ J/kg °C})(T_f - 35.0) = -(0.750 \text{ kg})(4186 \text{ J/kg °C})(T_f - 12.0) }\)
\(\mathsf{ (205.11)(T_f - 35.0) = -(3139.5)(T_f - 12.0) }\)
\(\mathsf{ 205.117T_f - 7178.85 = -3139.5T_f + 37674 }\)
\(\mathsf{ 205.11T_f + 3139.5T_f = 37674 + 7178.85 }\)
\(\mathsf{ 3344.61T_f = 44852.85 }\)
\(\mathsf{ T_f = \frac{44852.85}{3344.61} }\)
\(\mathsf{ T_f = 13.4 \text{°C} }\)
