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hammer [34]
3 years ago
7

A 40.5 g sample of aluminum is cooled 7.5°C. The specific heat capacity of aluminum is

Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
3 0
Q=mc\Delta T
Q= 40.5g x 0.900 J/g degree Celsius x 7.5
Q= 273.4 J.
Even if it is cooled , we use the same temperature as heating it.
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If the specific heat of a solution is 4.18 J/goC, and you have 296 mL (1.03 g/mL) which increases in temperature by 6.9 degrees,
Sergeu [11.5K]

Answer:

Q = 8.8 kJ

Explanation:

Step 1: Data given

The specific heat of a solution = 4.18 J/g°C

Volume = 296 mL

Density = 1.03 g/mL

The temperature increases with 6.9 °C

Step 2: Calculate the mass of the solution

mass = density * volume

mass = 1.03 g/mL * 296 mL

mass = 304.88 grams

Step 3: Calculate the heat

Q = m*c*ΔT

⇒ with Q = the heat in Joules = TO BE DETERMINED

⇒ with m = the mass of the solution = 304.88 grams

⇒ with c = the specific heat of the solution = 4.18 J/g°C

⇒ with ΔT = the change in temperature = 6.9 °C

Q = 304.88 g * 4.18 J/g°c * 6.9 °C

Q = 8793.3 J = 8.8 kJ

Q = 8.8 kJ

8 0
3 years ago
Why is salt a mineral
shutvik [7]
Salt is a mineral composed primarily of sodium chloride (NaCl), a chemical compound belonging to the larger class of salts; salt in the form of a natural crystalline mineral is known as rock salt or halite. ... Salt is essential for life in general, and saltiness is one of the basic human tastes
8 0
2 years ago
Read 2 more answers
If you observe a Full Moon on January 12th, on what date would you observe the next New Moon?
AlexFokin [52]

January 19

Explanation:

every first 7 days of a month the moon roughly starts its full moon phase

8 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
3 years ago
This is not a homework question
Dovator [93]
What state you live in? because I know someone who does it for free
8 0
3 years ago
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