Question

A metal object with mass of 29.0 g is heated to 97.0 °C and then transferred to an insulated container containing 95.8 g of water at 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 23.7 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.

Answer 1

0.603J/g°C is the specific heat of the metal object

Step-by-step explanation:

To solve this question we must know the heat given by the metal is equal to the heat absorbed by the water. The change in heat follows the equation:

Q = m*S*ΔT

Where Q is heat in Joules, m is the mass of the substance, S its specific heat and ΔT change in temperature

The equation to solve the problem is:

m(Object)*S(Object)*ΔT(Object) = m(Water)*S(Water)*ΔT(Water)

Where:

m(Object) = 29.0g

S(Object) = ??

ΔT(Object) = (97.0°C - 23.7°C = 73.3°C)

m(Water) = 95.8g

S(Water) = 4.184J/g°C

ΔT(Water) = (23.7°C - 20.5°C = 3.2°C)

29.0g*S(Object)*73.3°C = 95.8g*4.184J/g°C*3.2°C

S (Object) * 2125.7g°C = 1282.6J

S(Object) = 0.603J/g°C is the specific heat of the metal object