Question

a 70.0 g piece of metal at 80.0 ^ { circ } c is placed in 100.0 g of water at 22.0 ^ { circ } c contained in a calorimeter. the metal and water come to the same temperature at 24.6 ^ { circ } c.
how much heat did the metal give up to the water?
j
what is the specific heat of the metal?
j / g ^ { circ } c
regardless how careful one may be in conducting the experiment describe above, we can assume that some amount of heat was lost to the surroundings. fill in the blanks below that describe how the values would be different if no heat were lost to the surroundings.
the final temperature would be ^ { circ } c.
the computed amount of heat lost by the metal would then be than what was calculated in the above problem.
the computed specific of the metal would then be than what was calculated in the above problem.
therefore the actual specific heat is than what was calculated above.

Explanation:

Step1: Calculate heat gained by water

The formula for heat \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity (\(c_{water}=4.184\ J/g^{\circ}C\)), and \(\Delta T\) is temperature change.
\(m_{water}=100.0\ g\), \(\Delta T_{water}=24.6 - 22.0=2.6^{\circ}C\)
\(Q_{water}=m_{water}c_{water}\Delta T_{water}=100.0\times4.184\times2.6\)
\(Q_{water}=1087.84\ J\)

Step2: Calculate specific heat of metal

Since \(Q_{metal}=-Q_{water}\) (heat lost by metal = heat gained by water), \(m_{metal} = 70.0\ g\), \(\Delta T_{metal}=24.6 - 80.0=- 55.4^{\circ}C\)
Using \(Q = mc\Delta T\), \(c_{metal}=\frac{Q_{metal}}{m_{metal}\Delta T_{metal}}=\frac{- 1087.84}{70.0\times(-55.4)}\)
\(c_{metal}=\frac{- 1087.84}{-3878}=0.2805\ J/g^{\circ}C\)

Step3: Analyze if no heat lost to surroundings

• Final temperature: If no heat is lost to surroundings, more heat from metal goes to water. So final temperature \(T_f\) would be higher.
• Heat lost by metal: \(Q = mc\Delta T\), \(\Delta T_{metal}=T_f - T_{i,metal}\), \(\Delta T_{water}=T_f - T_{i,water}\). With higher \(T_f\), \(\Delta T_{water}\) is larger. Since \(Q_{water}=mc\Delta T\), \(Q_{water}\) (which equals \(Q_{metal}\)) is larger.
• Computed specific heat of metal: \(c_{metal}=\frac{Q_{metal}}{m_{metal}\Delta T_{metal}}\). With larger \(Q_{metal}\) (numerator) and same \(m_{metal}\) and \(\Delta T_{metal}\) (denominator), computed \(c_{metal}\) is larger. But since actual \(Q_{metal}\) (when no heat loss) is larger than assumed (when heat loss to surroundings), the calculated \(c_{metal}\) (with heat loss assumption) is smaller. So actual \(c_{metal}\) is larger.

Answer:

Heat given up: \(1087.84\ J\)
Specific heat of metal: \(0.2805\ J/g^{\circ}C\)
Final temperature: higher
Computed heat lost by metal: larger
Computed specific heat of metal: larger
Actual specific heat: larger