Question

A. How many Joules of energy are required to raise 3.50kg of water from 38.5 to 75.0C ? (specific heat of water is 4.184J/g C)(1000g=1kg)

b. How many grams of carbon are present if it requires 37.5 kJoules of energy to heat a sample from 30.0 to 55.0C ? (specific heat of carbon is 0.71 J/g C)

c. If 480.0 Joules is applies to a 50.0g sample of Hg in a thermometer that reads 25.0C, what will be the final temperature of the sample? (SH mercury is 0.14)

d. A 25.0kg sample of an unknown metal X requires 875kJ of energy to heat it from 55.0 to 125.0C. What is the specific heat of the unknown metal?


5.690x103 miles to meters.

900.0kg to grams

Please also give sig figs to all answers.

Answer fast please.

Answer 1

a.

A substance's specific heat tells you how much heat is required to increase the temperature of 1 g of that substance by 1°C.

The equation that establishes a relationshop between heat and change in temperature is

q = m • c • ∆T, where

q - heat absorbed

c - the specific heat of the substance, in your case of water

ΔT - the change in temperature, defined as the difference between the final temperature and the initial temperature

so:

q = 1.00 g • 4.18 J/g×°C • (75.0 - 38.5)°C

q = 152,57 J

just apply this formula for all exercises

Answer 2

A. How many Joules of energy are required to raise 3.50kg of water from 38.5 to 75.0C ? (specific heat of water is 4.184J/g C)(1000g=1kg)

Answer:

q = 534,506 Joules

q = 534.506 KiloJoules

534.506 KiloJoules or 534,506 Joules energy required to raise 3.50kg of water from 38.5 to 75.0C

What is sp.heat ?

Sp. Heat is the heat required to increase the temperature of the 1 mass of a given substance by a
`1^(0)` C temperature.

The formula of specific heat Cp =
`(q)/(m (∆T))`

Where,

q = energy of substance (Joules / KiloJoules),

Cp = Specific heat capacity of the substance (J/Kg.C),

m = mass of the substance

∆T = Change in temp.

Step-by-step explanation:

Given data from que: -

Mss of the water (m) = 3.50 Kg = 3.50 × 1000 gm = 3500 gm

specific heat capacity of water (Cp) = 4.184J/g C

Change in temp (∆T) = 75.0 - 38.5 = 36.5

now, put all above given data in formula

we get

Cp =
`(q)/(m (∆T))`

q = Cp ×m×∆T

q = 4.184×3500×36.5

q = 534,506 Joules

q = 534.506 KJ