A 400 g marble block has a specific heat of 860 J/kg C deg. If the block is placed in 700 ml of water, the marble changes it temperature by 17 Celsius degrees. Since the specific heat of water is known to be 4186 J/kg C deg, what is the temperature change of the water. (The block was cold and the water was warm before being allowed to equalize.)
To find the temperature change of the water, we can use the equation for heat transfer:
Q = mcΔT
Where:
Q is the heat transfer (in joules),
m is the mass of the object (in kg),
c is the specific heat capacity (in J/kg °C),
ΔT is the change in temperature (in °C).
In this case, we have the following information:
Mass of the marble block (m1) = 400 g = 0.4 kg
Specific heat of the marble block (c1) = 860 J/kg °C
Change in temperature of the marble block (ΔT1) = 17 °C
Mass of the water (m2) = 700 ml (which is equivalent to 700 g because the density of water is approximately 1 g/ml)
Specific heat of water (c2) = 4186 J/kg °C
Change in temperature of the water (ΔT2) = ?
Now, let's substitute the given values into the equation:
Q1 = m1 * c1 * ΔT1
Q2 = m2 * c2 * ΔT2
Since heat is transferred from the marble block to the water and we know that the total heat transferred (Q) is equal for both, we have:
Q1 = Q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Now, we can solve for ΔT2:
ΔT2 = (m1 * c1 * ΔT1) / (m2 * c2)
Plugging in the given values:
ΔT2 = (0.4 kg * 860 J/kg °C * 17 °C) / (0.7 kg * 4186 J/kg °C)
Calculating this equation will give us the temperature change of the water.