To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.50 J/g°C, if the initial temperature of the glass is 20.0°C?

Answer 1

#230^@"C"#

A substance's specific heat tells you how much heat much either be added or removed from #"1 g"# of that substance in order to cause a #1^@"C"# change in temperature.
The change in temperature, #DeltaT#, is always calculated by subtracting the initial temperature of the sample from the final temperature of the sample.
#color(blue)(DeltaT = T_"final" - T_"initial")#
Now, when the substance absorbs heat, its temperature will increase, which implies that #DeltaT > 0#.

The objective here is to first identify the temperature change and then utilize that information to determine the sample's final temperature.

You must apply this formula.

#color(blue)(q = m * c * DeltaT)" "#, where
#q# - the amount of heat added / removed #m# - the mass of the sample #c# - the specific heat of the substance #DeltaT# - the change in temperature

This equation, as you can see, establishes a relationship between the mass of a substance, its specific heat, the amount of heat added or removed from the sample, and the temperature change that results.

In your case, adding #"5275 J"# of heat to that #"50.0-g"# piece of glass will result in a temperature change of
#q = m * c * DeltaT implies DeltaT = q/(m * c)#

Enter your values to obtain

#DeltaT = (5275 color(red)(cancel(color(black)("J"))))/(50.0color(red)(cancel(color(black)("g"))) * 0.50color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 211^@"C"#
So, adding that much heat to your sample will result in a #211^@"C"# increase in temperature. This means that the final temperature of the glass will be
#T_"final" = T_"initial" + DeltaT#
#T_"final" = 20.0^@"C" + 211^@"C" = color(green)(230^@"C")#

The number of sig figs you have for the particular heat of glass is the answer, which is rounded to two sig figs.

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Answer 2

( \Delta T = \frac{q}{m \cdot c} ) ( \Delta T = \frac{5275 , \text{J}}{50.0 , \text{g} \cdot 0.50 , \text{J/g°C}} ) ( \Delta T = 211 , \text{°C} ) Final temperature: ( 20.0 , \text{°C} + 211 , \text{°C} = 231 , \text{°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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