An object with a mass of #20 g# is dropped into #120 mL# of water at #0^@C#. If the object cools by #60 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

Answer 1

#1260 "J kg"^{-1}"K"^{-1}#

The specific heat capacity of water is #4200 "J kg"^{-1}"K"^{-1}#. So, the heat capacity of 120 mL = 0.12 kg water is #504 "J K"^{-1}#. Thus the heat gained by the water is 1512 J. Assuming that no energy is lost in the process, this is the heat lost by the body in cooling down by 60 K. Thus the specific heat capacity of the material is given by
#{1512 "J"}/{0.02 "kg" xx 60 "K"} = 1260 "J kg"^{-1}"K"^{-1}#
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Answer 2

Specific heat of the material = (\frac{{m \times c \times \Delta T_{\text{object}}}}{{m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}}}}) Where: (m) = mass of the object (in grams) (c) = specific heat capacity of the material (in J/g°C) (\Delta T_{\text{object}}) = change in temperature of the object (in °C) (m_{\text{water}}) = mass of water (in grams) (c_{\text{water}}) = specific heat capacity of water (in J/g°C) (\Delta T_{\text{water}}) = change in temperature of water (in °C)

Given: (m = 20 \text{ g}) (\Delta T_{\text{object}} = -60 \text{ °C}) (negative because it cools) (m_{\text{water}} = 120 \text{ g}) (\Delta T_{\text{water}} = 3 \text{ °C}) (c_{\text{water}} = 4.18 \text{ J/g°C}) (specific heat capacity of water)

(c = \frac{{m \times c \times \Delta T_{\text{object}}}}{{m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}}}}) (c = \frac{{20 \times c \times (-60)}}{{120 \times 4.18 \times 3}}) (c = \frac{{-1200 \times c}}{{1500.6}}) (c = \frac{{-1200}}{{1500.6 \times c}}) (c = \frac{{-1200}}{{1500.6}}) (c ≈ -0.799 \text{ J/g°C})

Therefore, the specific heat of the material that the object is made of is approximately (0.799 \text{ J/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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