1) Nova, whose mass is 50.0 kg, stays out skiing for too long and her body temperature drops by 2.00 degree celsius . What is the amount of heat lost from Nova's body? (cHuman body= 3470 J/kg degre celcius)
2) Phoebe's insulated foam cup is filled with 0.150 kg of the coffee (mostly water)that is too to hot to drink, so she adds 0.010 kg of milk at 5.0 degree celcius. If the coffee has an initial of temperature of 70.0 degree celcius and the specific heat of milk is 3800 J/kg degree celcius, how hot is the coffee after the milk is added? (Assume that no heat leaks out through the cup)
a. heat=mc*deltaTemp use c for water
b. Sum of heats gained is zero.
heatgainedmilk+heatgainedcoffee=0
.01*cmilk*(Tf-5)+.150*cw*(Tf-70)=0
solve for Tf, given cmilk, and cwater.
To solve these problems, we will be using the formula:
Q = mcΔT
Where:
Q is the heat energy transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Let's calculate the amount of heat lost from Nova's body first.
Given:
Mass of Nova (m) = 50.0 kg
Change in temperature (ΔT) = -2.00 °C
Specific heat capacity of a human body (c) = 3470 J/kg°C
Using the formula Q = mcΔT, we can substitute the given values:
Q = (50.0 kg)(3470 J/kg°C)(-2.00 °C)
Solving this equation, we get:
Q = -347,000 J
However, we usually represent the energy lost as a positive value. Thus, the amount of heat lost from Nova's body is 347,000 J.
Now, let's move on to the second problem.
Given:
Mass of coffee (m_coffee) = 0.150 kg
Initial temperature of coffee (T_initial) = 70.0 °C
Mass of milk (m_milk) = 0.010 kg
Temperature of milk (T_milk) = 5.0 °C
Specific heat capacity of milk (c_milk) = 3800 J/kg°C
We need to find the final temperature of the coffee after the milk is added.
To find the heat gained by the coffee, we need to calculate the heat gained by each substance separately and then combine them.
Heat gained by coffee (Q_coffee) = mcΔT
Q_coffee = (0.150 kg)(4200 J/kg°C)(T_final - 70.0 °C)
Heat gained by milk (Q_milk) = mcΔT
Q_milk = (0.010 kg)(3800 J/kg°C)(T_final - 5.0 °C)
Since no heat leaks out through the cup, the total heat gained by the coffee and milk should be equal to zero:
Q_coffee + Q_milk = 0
Substituting the calculated values:
(0.150 kg)(4200 J/kg°C)(T_final - 70.0 °C) + (0.010 kg)(3800 J/kg°C)(T_final - 5.0 °C) = 0
Simplifying the equation:
63T_final - 4410 + 38T_final - 190 = 0
Combining like terms:
101T_final - 4600 = 0
Solving for T_final:
T_final = 4600 / 101
T_final ≈ 45.5 °C
Therefore, the coffee's final temperature after the milk is added will be approximately 45.5 °C.
To solve the given problems, we will use the formula:
Q = mcΔT
where:
Q represents the heat lost or gained
m represents the mass of the substance
c represents the specific heat capacity of the substance
ΔT represents the change in temperature
Let's solve each problem step by step:
1) Nova's heat loss:
We are given:
Mass (m) = 50.0 kg
Change in temperature (ΔT) = -2.00 °C
Specific heat capacity (c) = 3470 J/kg °C
Plugging these values into the formula:
Q = (m)(c)(ΔT)
= (50.0 kg)(3470 J/kg °C)(-2.00 °C)
= -347,000 J
The negative sign indicates the heat is lost from Nova's body. Therefore, the amount of heat lost from Nova's body is 347,000 J.
2) Heating the coffee:
We are given:
Mass of coffee (m) = 0.150 kg
Initial temperature of coffee (ΔT) = 70.0 °C
Final temperature of coffee = ?
Specific heat capacity of coffee (c) = unknown
We need to find the specific heat capacity of coffee, so we will use the formula:
Q = mcΔT
The heat gained by the coffee will be equal to the heat lost by the milk. So:
Qcoffee = Qmilk
(m)(c)(ΔT) = (m)(c)(ΔT)
Plugging in the values we know:
(mcoffee)(ccoffee)(ΔTcoffee) = (mmilk)(cmilk)(ΔTmilk)
(0.150 kg)(ccoffee)(70.0 °C - ΔTcoffee) = (0.010 kg)(3800 J/kg °C)(70.0 °C - 5.0 °C)
Now we can solve for cc:
ccoffee = [(0.010 kg)(3800 J/kg °C)(70.0 °C - 5.0 °C)] / [(0.150 kg)(70.0 °C - ΔTcoffee)]
ccoffee = 2533.33 J/kg °C / (70.0 °C - ΔTcoffee)
Next, we rearrange the equation to solve for the final temperature of coffee (ΔTcoffee):
ccoffee = 2533.33 J/kg °C / (70.0 °C - ΔTcoffee)
ccoffee(70.0 °C - ΔTcoffee) = 2533.33 J/kg °C
70.0 °C - ΔTcoffee = 2533.33 J/kg °C / ccoffee
ΔTcoffee = 70.0 °C - (2533.33 J/kg °C / ccoffee)
Now, we need to calculate the specific heat capacity of coffee (ccoffee):
ccoffee = (0.010 kg)(3800 J/kg °C) / (70.0 °C - ΔTcoffee)
Once we have the value of ccoffee, we substitute it back into the equation to find ΔTcoffee.
Using these equations, you can calculate the final temperature of the coffee after the milk is added.