A 920 g empty iron kettle is put on a stove. How much heat in joules must it absorb to raise its temperature from 15.0° C to 93.0° C? (The specific heat for iron is 113 cal/kg · °C.)(1 cal = 4.184 J)

q=Mass*specific heat*change in temp=(0.920kg)(113cal/kg *C)(78 C). Take answer for that you get and multiply it by 4.184J/cal.

Well, it sounds like that kettle is in hot water! Let's start by calculating the mass of the kettle in kilograms. To do that, we divide the mass in grams by 1000:

920 g ÷ 1000 = 0.92 kg

Now, we'll use the formula for heat energy:

Q = m * c * ΔT

where Q is the heat energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, let's convert the specific heat from cal/(kg * °C) to J/(kg * °C):

113 cal/(kg * °C) × (1 J ÷ 4.184 cal) = 27.084 J/(kg * °C)

Now, let's plug in all the values and calculate Q:

Q = (0.92 kg) * (27.084 J/(kg * °C)) * (93.0°C - 15.0°C)

To calculate the heat absorbed by the iron kettle, we can use the formula:

Q = m * c * ΔT

Where:
Q = Heat absorbed (in joules)
m = Mass of the kettle (in kg)
c = Specific heat capacity of iron (in cal/kg · °C)
ΔT = Change in temperature (in °C)

First, we need to convert the mass of the kettle from grams to kilograms:

m = 920 g / 1000 = 0.92 kg

Next, we can convert the specific heat capacity from calories to joules:

c = 113 cal/kg · °C * 4.184 J/cal = 472.192 J/kg · °C

Now, we can calculate the change in temperature:

ΔT = 93.0°C - 15.0°C = 78.0°C

Finally, we can substitute the values into the formula to find the heat absorbed:

Q = 0.92 kg * 472.192 J/kg · °C * 78.0°C
Q = 33,713.9856 J

Therefore, the iron kettle must absorb approximately 33,713.99 Joules of heat to raise its temperature from 15.0°C to 93.0°C.

To calculate the amount of heat required to raise the temperature of the iron kettle, we can use the formula:

Q = mcΔT

Where:
Q is the heat absorbed (in joules),
m is the mass of the kettle (in kilograms),
c is the specific heat capacity of iron (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degree Celsius).

First, let's convert the mass of the kettle from grams to kilograms:
920 g = 920/1000 kg = 0.92 kg

Next, let's convert the specific heat capacity from cal/kg · °C to joules per kilogram per degree Celsius:
1 cal = 4.184 J

Therefore, the specific heat capacity of iron is:
113 cal/kg · °C x 4.184 J/cal = 472.792 J/kg · °C (rounded to 3 decimal places)

Now, we can substitute the values into the formula:

Q = (0.92 kg) x (472.792 J/kg · °C) x (93.0°C - 15.0°C)

Q = 0.92 kg x 472.792 J/kg · °C x 78.0°C

Q ≈ 33,438.678 J

Therefore, the kettle must absorb approximately 33,438.678 joules of heat to raise its temperature from 15.0°C to 93.0°C.