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2 years ago

If 0.092J of heat causes a 0.267 degree C temperature change, what mass of water is present?

Chemistry

1 answer:

Thepotemich [5.8K]2 years ago

6 0

Answer:

0.082g

Explanation:

The following data were obtained from the question:

Heat (Q) = 0.092J

Change in temperature (ΔT) = 0.267°C

Specific heat capacity (C) of water = 4.184J/g°C

Mass (M) =..?

Thus, the mass of present can be obtained as follow:

Q = MCΔT

0.092 = M x 4.184 x 0.267

Divide both side by 4.184 x 0.267

M = 0.092 / (4.184 x 0.267)

M = 0.082g

Therefore, mass of water was present is 0.082.

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

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Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

The flame in a torch used to cut metal is produced by burning acetylene (C2H2) in pure oxygen. Assuming the combustion of 1 mole

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Answer:

19.8 kg of C₂H₂ is needed

Explanation:

We solve this by a rule of three:

If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene

95.5×10⁴ kJ of heat may be released by the combustion of

(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂

Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g

If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg

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2 years ago

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