How much heat is added when 22 grams of water are heated from 10°C to 15°C?

Answer 1

And the heat capacity of water is #??# #J*K^-1*g^-1#?

We have #"mass"#, we have #DeltaT#. We don't have no heat capacity of water.

SO I will take pity on you. The specific heat capacity of water is #4.184*J*K^-1*g^-1#. As specific heats go, this is very large.

And thus #DeltaH# #=# #"Mass "xx" Specific heat "xx""DeltaT#

#=# #22*gxx4.184*J*(""^@C)^-1*g^-1xx5""^@C# #~=500*J#

I am free to interchange #"degrees Kelvin"# and #"degrees Celsius"# because #DeltaT# is specified.

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Answer 2

Kai Bowman

The amount of heat added can be calculated using the formula:

Q = mcΔT

Where: Q = amount of heat added (in joules) m = mass of the substance (in grams) c = specific heat capacity of the substance (in J/g°C) ΔT = change in temperature (in °C)

For water, the specific heat capacity (c) is 4.18 J/g°C.

Given: m = 22 grams ΔT = 15°C - 10°C = 5°C

Plugging in the values:

Q = (22 g) * (4.18 J/g°C) * (5°C) Q = 462.2 joules

Therefore, 462.2 joules of heat are added when 22 grams of water are heated from 10°C to 15°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.