How would you calculate the specific beat of silver, given that 280 g of the metal absorbed 136 calories when changed from 22.13°C to 30.79°C?

Answer 1

#c_"Ag"=0.056"cal"/("g"*^@"C")#

The equation you'll need is the following:
, #m# is mass in grams, and #DeltaT# is the change in Celcius temperature. #DeltaT=T_"final"-T_"initial"#

#Q="136 cal"#
#m="280 g"#
#T_"initial"="22.13"^@"C"#
#T_"final"="30.79"^@"C"#
#DeltaT="30.79"^@"C"-"22.13"^@"C"#"="8.66"^@"C"#

Rearrange the equation to isolate #c#.

#c_"Ag"=Q/(mDeltaT)#

#c_"Ag"=(136"cal")/(280"g"*8.66^@"C")#

#c_"Ag"=0.056"cal"/("g"*^@"C")#

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Answer 2

To calculate the specific heat of silver, you can use the formula:

[ q = mc\Delta T ]

Where:

  • ( q ) = heat absorbed (in calories)
  • ( m ) = mass of the substance (in grams)
  • ( c ) = specific heat capacity (in calories/gram °C)
  • ( \Delta T ) = change in temperature (in °C)

Rearrange the formula to solve for specific heat (( c )):

[ c = \frac{q}{m\Delta T} ]

Substitute the given values:

[ c = \frac{136 \text{ cal}}{280 \text{ g} \times (30.79°C - 22.13°C)} ]

Calculate:

[ c ≈ \frac{136 \text{ cal}}{280 \text{ g} \times 8.66°C} ≈ \frac{136 \text{ cal}}{2420.8 \text{ cal}} ]

[ c ≈ 0.0562 \text{ cal/g°C} ]

So, the specific heat of silver is approximately ( 0.0562 \text{ cal/g°C} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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