**Purpose:** To find the specific heat capacity of a substance

**Principle: ** According to the principle of thermal energy exchange, when thermal energy is transferred from a warmer object to a colder object, the amount of thermal energy released by the warmer object is equal to the amount of thermal energy absorbed by the colder object.

Q _{released} =Q_{absorbed}

The quantity of heat required to heat up a substance through T is Q = mc∆T

m_{1} c_{1 }∆T_{1 }= m_{2 }c_{2} ∆T_{2} ( or ) c_{1}= (m_{2 }c_{2} ∆T_{2} ) / m_{1} ∆T_{1}

**Procedure:**

Measure the mass of water

Measure the initial temperatures of water and cube

Drop the cube in water and note the final steady temperature of the system

Measure the mass of the cube

Repeat the experiment with different cubes

Calculate the temperature changes that happen to the water and the cube

Calculate the specific heat capacity of each cube

**Observations:**

Mass of water m_{2}= **224g = 0.224 kg**

Specific heat capacity of water c_{2} = **4200 J / kg ^{}C**

Initial Temp of water ( t_{2 }) ^{}C | Initial Temp of cube ( t_{1 }) ^{ 0}C | Final Temp of the mixture ( T ) ^{}C | Mass of cube ( m_{1 }) kg | |

Cube 1 | 22.5^{}C | -6^{}C | 18^{}C | 0.205 kg |

Cube 2 | 20.5^{}C | -6^{}C | 20^{}C | 0.066 kg |

Cube 3 | 21^{}C | -6^{}C | 20^{}C | 0.068 kg |

**Calculations:**

∆T_{2 }= t_{2} – T ( ^{}C ) ( Water ) | ∆T_{1} = T – t_{1} ( ^{}C ) ( Cube ) | c_{1} = (m_{2 }c_{2} ∆T_{2} ) / m_{1} ∆T_{1} J / kg ^{}C | |

Cube 1 | 22.5 – 18 = 4.5^{}C | 18 – (–6) = 24^{}C | (0.224)(4200)(4.5)/(0.205)(24) = 861.6 J/kg℃ |

Cube 2 | 20.5 – 20 = 0.5^{}C | 20 – (–6) = 26^{}C | (0.224)(4200)(0.5)/(0.066)(26) = 274.1 J/kg℃ |

Cube 3 | 21 – 20 = 1^{}C | 20 – (–6) = 26^{}C | (0.224)(4200)(1)/(0.068)(26) = 532 J/kg℃ |

The calculated value of specific heat capacity of cube 1 = **861.6 J/kg℃**

The calculated value of specific heat capacity of cube 2 = **274.1 J/kg℃**

The calculated value of specific heat capacity of cube 3 = **532 J/kg℃**

Actual Specific heat capacity of cube 1 = **880 J/Kg℃** (Based on Researched heat capacity for Aluminum)

Actual Specific heat capacity of cube 2 = **236** **J/Kg℃ **(Based on Researched heat capacity for Silver)

Actual Specific heat capacity of cube 3 = **521** **J/Kg℃** (Based on Researched heat capacity for Titanium)

% error for cube **1** = (Observed – Actual)/Actual * 100% = (861.6-880)/880 *100% = **2.09 %**

% error for cube **2** = (Observed – Actual)/Actual * 100% = (274.1-236)/236 *100% = **16.14 **%

% error for cube **3** = (Observed – Actual)/Actual * 100% = (532-521)/521 *100% = **2.11 %**

**Conclusion:**

In conclusion, the purpose of this lab was to determine the specific heat capacity of a substance by using the principle of thermal energy exchange and the equation Q = mc∆T. By measuring the temperature changes and masses of the substance and water in three separate experiments, we were able to calculate the specific heat capacity of the substance using the equation m1c1∆T1 = m2c2∆T2. The accuracy of our results may have been affected by various sources of error, such as heat loss to the surroundings and the assumption of a perfectly insulated system. Nonetheless, this lab provided a valuable opportunity to apply fundamental principles of thermodynamics and to gain practical experience with experimental techniques for measuring thermal properties.