Question

Calculate the amount of heat needed to increase the temperature of 200 grams of water at 10°C to 95°C.

Answer 1

the amount of heat need is 71060 Joules

Explanation

The equation for the amount of heat needed to increase the temperature of water is as follows:

$\begin{gathered} E=mC(T_2-T_1) \\ where \\ \\ \end{gathered}$

where E is the amount of heat

C is the specific heat capacity of water having a standard value of 4180 J/kg*K

T is the final ans initial temperature in kelvin

m is the mass

so

Step 1

given

$\begin{gathered} mass=m=200\text{ gr=0.2 kg} \\ T_1=10\text{ \degree C} \\ T_2=95\text{ \degree C} \\ C=4180\text{ J/kh*K} \end{gathered}$

a) convert the temperature into kelvin

$\begin{gathered} 10\text{ \degree C=\lparen10+273\rparen K=283 K} \\ 95\text{ \degree C=\lparen95+273 \rparen K=368 K} \end{gathered}$

now, replace in the formula

$\begin{gathered} E=mC(T_{2}-T_{1}) \\ E=0.2\text{ kg*4180L/kg*K*\lparen368K-283 k\rparen} \\ E=71060\text{ J} \end{gathered}$

therefore, the amount of heat need is 71060 Joules

I hope this helps you

Answer 2

71.06 kJ

Heat is the form of energy that is produced by combustion of fuel. The heat will flow due to temperature gradient without any external means. the temperature of any system increases due to heat energy.

coming to the numerical portion of question: -

mass of water=200 grams (since grams is not the standard international unit of weight so we will convert grams into kilograms)

1 gram = 0.001 kg

So, applying unitary method: -

200 grams = 200 x 0.001 = 0.2 kg

initial temperature of water= 10-degree Celsius

since degree Celsius is not the SI unit of temperature we will convert Celsius to kelvin

kelvin temperature = degree Celsius + 273

therefore 10-degree Celsius = 10+273=283 k

similarly final temperature of water = 95+273=368k

now using equation for amount of heat needed to increase the temperature: -

E=mC(final temperature - initial temperature)  

here C is the specific heat capacity of water having standard value of 4180 J/kg

substituting the values

E=(0.2 kg)(4180 J/kg .k)(368k-283k)

E=71060J x (1kJ/1000J)

E= 71.06 KJ

For more information about heat refer the below mentioned link :-

brainly.com/question/13411214