4. A 1.00 × 10² g aluminum block at 100.0℃ is placed in 1.00 × 10² g of water at 10.0℃. The final temperature of the mixture is 26.0℃. What is the specific heat of the aluminum?
5. Three metal fishing weights, each with a mass of 1.00 × 10² g and at a temperature of 100.0℃, are placed in 1.00 × 10² g of water at 35.0℃. The final temperature of the mixture is 45.0℃. What is the specific heat of the metal in the weights?
SOLUTION :
4.
Heat flows from higher temperature object to lower temperature object when they come
In contact. So, heat is transferred from aluminium at 100ºC to water at 10ºC.
As per energy system,
Heat given by aluminium = Heat taken by water.
=> m_{a} * s_{a} * (t_{i} - t_{f}) = m_{w} * s_{w} * (_{f}f - t_{i})
=> 1.00*10^2 * s_{a} * (100 - 26) = 1*10^2 * 1.00 * (26 - 10)
=> 74.00*10^2 s_{a} = 16.00*10^2
=> s_{a} = 16.00/74.00 = 0.2162 cal/(g*ºC)
So, specific heat of aluminium is 0.2162 cal/g ºC (ANSWER)
5.
Heat flows from higher temperature object to lower temperature object when they come
In contact. So, heat is transferred from metal fishing weight at 100ºC to water at 35ºC.
As per energy system,
Heat given by metal fishing weight = Heat taken by water.
=> m_{m} * s_{m} * (t_{i} - t_{f}) = m_{w} * s_{w} * (t_{f} - t_{i})
=> 1.00*10^2 * s_{m} * (100 - 45) = 1*10^2 * 1.00 * (45 - 35)
=> 55.00*10^2 s_{m} = 10.00*10^2
=> s_{m} = 10.00/55.00 = 0.1818 cal/(g*ºC)
So, specific heat of metal fishing weight is 0.1818 cal/g ºC (ANSWER)
4. specific heat of aluminium is 0.2162 cal/g ºC
5. specific heat of metal fishing weight is 0.1818 cal/g ºC
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